Class 9: Expansion – Exercise 4

$\displaystyle \textbf{Exercise 4(A)}$

$\displaystyle \textbf{Question 1: }\text{Find the square of:}$
$\displaystyle \text{(i) }2a+b \qquad \text{(ii) }3a+7b \qquad \text{(iii) }3a-4b \qquad \text{(iv) }\frac{3a}{2b}-\frac{2b}{3a}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }\ (2a+b)^2$
$\displaystyle =(2a)^2+2(2a)(b)+b^2$
$\displaystyle =4a^2+4ab+b^2$
$\displaystyle \therefore (2a+b)^2=4a^2+4ab+b^2.$

$\displaystyle \text{(ii) }\ (3a+7b)^2$
$\displaystyle =(3a)^2+2(3a)(7b)+(7b)^2$
$\displaystyle =9a^2+42ab+49b^2$
$\displaystyle \therefore (3a+7b)^2=9a^2+42ab+49b^2.$

$\displaystyle \text{(iii) }\ (3a-4b)^2$
$\displaystyle =(3a)^2-2(3a)(4b)+(4b)^2$
$\displaystyle =9a^2-24ab+16b^2$
$\displaystyle \therefore (3a-4b)^2=9a^2-24ab+16b^2.$
$\displaystyle \\$

$\displaystyle \text{(iv) }\ \left(\frac{3a}{2b}-\frac{2b}{3a}\right)^2$
$\displaystyle =\left(\frac{3a}{2b}\right)^2-2\left(\frac{3a}{2b}\right)\left(\frac{2b}{3a}\right)+\left(\frac{2b}{3a}\right)^2$
$\displaystyle =\frac{9a^2}{4b^2}-2+\frac{4b^2}{9a^2}$
$\displaystyle \therefore \left(\frac{3a}{2b}-\frac{2b}{3a}\right)^2=\frac{9a^2}{4b^2}-2+\frac{4b^2}{9a^2}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{Use identities to evaluate:}$
$\displaystyle \text{(i) }(101)^2 \qquad \text{(ii) }(502)^2 \qquad \text{(iii) }(97)^2 \qquad \text{(iv) }(998)^2$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }\ (101)^2=(100+1)^2$
$\displaystyle =100^2+2(100)(1)+1^2$
$\displaystyle =10,000+200+1$
$\displaystyle =10,201$
$\displaystyle \therefore (101)^2=10,201.$

$\displaystyle \text{(ii) }\ (502)^2=(500+2)^2$
$\displaystyle =500^2+2(500)(2)+2^2$
$\displaystyle =250,000+2,000+4$
$\displaystyle =252,004$
$\displaystyle \therefore (502)^2=252,004.$

$\displaystyle \text{(iii) }\ (97)^2=(100-3)^2$
$\displaystyle =100^2-2(100)(3)+3^2$
$\displaystyle =10,000-600+9$
$\displaystyle =9,409$
$\displaystyle \therefore (97)^2=9,409.$

$\displaystyle \text{(iv) }\ (998)^2=(1000-2)^2$
$\displaystyle =1000^2-2(1000)(2)+2^2$
$\displaystyle =10,00,000-4,000+4$
$\displaystyle =996,004$
$\displaystyle \therefore (998)^2=996,004.$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{Evaluate:}$
$\displaystyle \text{(i) }\left(\frac78x+\frac45y\right)^2 \qquad \text{(ii) }\left(\frac{2x}{7}-\frac{7y}{4}\right)^2$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }\left(\frac78x+\frac45y\right)^2$
$\displaystyle =\left(\frac78x\right)^2+2\left(\frac78x\right)\left(\frac45y\right)+\left(\frac45y\right)^2$
$\displaystyle =\frac{49x^2}{64}+\frac{7xy}{5}+\frac{16y^2}{25}$
$\displaystyle \therefore \left(\frac78x+\frac45y\right)^2=\frac{49x^2}{64}+\frac{7xy}{5}+\frac{16y^2}{25}.$

$\displaystyle \text{(ii) }\left(\frac{2x}{7}-\frac{7y}{4}\right)^2$
$\displaystyle =\left(\frac{2x}{7}\right)^2-2\left(\frac{2x}{7}\right)\left(\frac{7y}{4}\right)+\left(\frac{7y}{4}\right)^2$
$\displaystyle =\frac{4x^2}{49}-xy+\frac{49y^2}{16}$
$\displaystyle \therefore \left(\frac{2x}{7}-\frac{7y}{4}\right)^2=\frac{4x^2}{49}-xy+\frac{49y^2}{16}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{Evaluate:}$
$\displaystyle \text{(i) }\left(\frac{a}{2b}+\frac{2b}{a}\right)^2-\left(\frac{a}{2b}-\frac{2b}{a}\right)^2-4$
$\displaystyle \text{(ii) }(4a+3b)^2-(4a-3b)^2+48ab$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) Let }x=\frac{a}{2b}\text{ and }y=\frac{2b}{a}.$
$\displaystyle \text{Then }(x+y)^2-(x-y)^2=4xy$
$\displaystyle \therefore \left(\frac{a}{2b}+\frac{2b}{a}\right)^2-\left(\frac{a}{2b}-\frac{2b}{a}\right)^2$
$\displaystyle =4\left(\frac{a}{2b}\right)\left(\frac{2b}{a}\right)$
$\displaystyle =4$
$\displaystyle \therefore \left(\frac{a}{2b}+\frac{2b}{a}\right)^2-\left(\frac{a}{2b}-\frac{2b}{a}\right)^2-4$
$\displaystyle =4-4$
$\displaystyle =0$
$\displaystyle \therefore \text{The required value is }0.$

$\displaystyle \text{(ii) }(4a+3b)^2-(4a-3b)^2+48ab$
$\displaystyle =(16a^2+24ab+9b^2)-(16a^2-24ab+9b^2)+48ab$
$\displaystyle =16a^2+24ab+9b^2-16a^2+24ab-9b^2+48ab$
$\displaystyle =96ab$
$\displaystyle \therefore \text{The required value is }96ab.$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{If }a+b=7\text{ and }ab=10,\text{ find }a-b.$
$\displaystyle \text{Answer:}$
$\displaystyle (a-b)^2=(a+b)^2-4ab$
$\displaystyle =(7)^2-4(10)$
$\displaystyle =49-40$
$\displaystyle =9$
$\displaystyle \therefore a-b=\pm3$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\text{If }a-b=7\text{ and }ab=18,\text{ find }a+b.$
$\displaystyle \text{Answer:}$
$\displaystyle (a+b)^2=(a-b)^2+4ab$
$\displaystyle =(7)^2+4(18)$
$\displaystyle =49+72$
$\displaystyle =121$
$\displaystyle \therefore a+b=\pm11$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: }\text{If }x+y=\frac{7}{2}\text{ and }xy=\frac{5}{2},\text{ find:}$
$\displaystyle \text{(i) }x-y \qquad \text{(ii) }x^2-y^2$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }(x-y)^2=(x+y)^2-4xy$
$\displaystyle =\left(\frac{7}{2}\right)^2-4\left(\frac{5}{2}\right)$
$\displaystyle =\frac{49}{4}-10$
$\displaystyle =\frac{49-40}{4}$
$\displaystyle =\frac{9}{4}$
$\displaystyle \therefore x-y=\pm\frac{3}{2}$

$\displaystyle \text{(ii) }x^2-y^2=(x+y)(x-y)$
$\displaystyle =\frac{7}{2}\left(\pm\frac{3}{2}\right)$
$\displaystyle =\pm\frac{21}{4}$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: }\text{If }a-b=0.9\text{ and }ab=0.36,\text{ find:}$
$\displaystyle \text{(i) }a+b \qquad \text{(ii) }a^2-b^2$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }(a+b)^2=(a-b)^2+4ab$
$\displaystyle =(0.9)^2+4(0.36)$
$\displaystyle =0.81+1.44$
$\displaystyle =2.25$
$\displaystyle \therefore a+b=\pm1.5$

$\displaystyle \text{(ii) }a^2-b^2=(a+b)(a-b)$
$\displaystyle =(\pm1.5)(0.9)$
$\displaystyle =\pm1.35$
$\displaystyle \\$

$\displaystyle \textbf{Question 9: }\text{If }a-b=4\text{ and }a+b=6,\text{ find:}$
$\displaystyle \text{(i) }a^2+b^2 \qquad \text{(ii) }ab$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }(a+b)^2=a^2+2ab+b^2$
$\displaystyle 6^2=a^2+2ab+b^2$
$\displaystyle 36=a^2+b^2+2ab\qquad\ldots(1)$
$\displaystyle (a-b)^2=a^2-2ab+b^2$
$\displaystyle 4^2=a^2-2ab+b^2$
$\displaystyle 16=a^2+b^2-2ab\qquad\ldots(2)$
$\displaystyle \text{Adding (1) and (2),}$
$\displaystyle 52=2(a^2+b^2)$
$\displaystyle \therefore a^2+b^2=26$

$\displaystyle \text{(ii) From (1),}$
$\displaystyle 36=26+2ab$
$\displaystyle 2ab=10$
$\displaystyle \therefore ab=5$
$\displaystyle \\$

$\displaystyle \textbf{Question 10: }\text{If }a+\frac1a=6\text{ and }a\ne0,\text{ find:}$
$\displaystyle \text{(i) }a-\frac1a \qquad \text{(ii) }a^2-\frac1{a^2}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }\left(a+\frac1a\right)^2=a^2+\frac1{a^2}+2$
$\displaystyle 6^2=a^2+\frac1{a^2}+2$
$\displaystyle \therefore a^2+\frac1{a^2}=34$
$\displaystyle \left(a-\frac1a\right)^2=a^2+\frac1{a^2}-2$
$\displaystyle =34-2$
$\displaystyle =32$
$\displaystyle \therefore a-\frac1a=\pm4\sqrt2$

$\displaystyle \text{(ii) }a^2-\frac1{a^2}=\left(a+\frac1a\right)\left(a-\frac1a\right)$
$\displaystyle =6\left(\pm4\sqrt2\right)$
$\displaystyle =\pm24\sqrt2$
$\displaystyle \\$

$\displaystyle \textbf{Question 11: }\text{If }a-\frac1a=8\text{ and }a\ne0,\text{ find:}$
$\displaystyle \text{(i) }a+\frac1a \qquad \text{(ii) }a^2+\frac1{a^2}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }\left(a-\frac1a\right)^2=a^2+\frac1{a^2}-2$
$\displaystyle 8^2=a^2+\frac1{a^2}-2$
$\displaystyle \therefore a^2+\frac1{a^2}=66$
$\displaystyle \left(a+\frac1a\right)^2=a^2+\frac1{a^2}+2$
$\displaystyle =66+2$
$\displaystyle =68$
$\displaystyle \therefore a+\frac1a=\pm2\sqrt{17}$

$\displaystyle \text{(ii) From above,}$
$\displaystyle a^2+\frac1{a^2}=66$
$\displaystyle \therefore \text{The required value is }66.$
$\displaystyle \\$

$\displaystyle \textbf{Question 12: }\text{If }a^2-3a+1=0\text{ and }a\ne0,\text{ find:}$
$\displaystyle \text{(i) }a+\frac{1}{a}\qquad \text{(ii) }a^2+\frac{1}{a^2}$
$\displaystyle \text{Answer:}$
$\displaystyle a^2-3a+1=0$
$\displaystyle \text{Dividing each term by }a,\text{ we get}$
$\displaystyle a-3+\frac{1}{a}=0$
$\displaystyle \text{(i) }\therefore a+\frac{1}{a}=3$

$\displaystyle \text{(ii) }a^2+\frac{1}{a^2}=\left(a+\frac{1}{a}\right)^2-2$
$\displaystyle =3^2-2$
$\displaystyle =9-2$
$\displaystyle =7$
$\displaystyle \therefore a^2+\frac{1}{a^2}=7$
$\displaystyle \\$

$\displaystyle \textbf{Question 13: }\text{If }a^2-5a-1=0\text{ and }a\ne0,\text{ find:}$
$\displaystyle \text{(i) }a-\frac{1}{a}\qquad \text{(ii) }a+\frac{1}{a}\qquad \text{(iii) }a^2-\frac{1}{a^2}$
$\displaystyle \text{Answer:}$
$\displaystyle a^2-5a-1=0$
$\displaystyle \text{Dividing each term by }a,\text{ we get}$
$\displaystyle a-5-\frac{1}{a}=0$
$\displaystyle \text{(i) }\therefore a-\frac{1}{a}=5$

$\displaystyle \text{(ii) }\left(a+\frac{1}{a}\right)^2=\left(a-\frac{1}{a}\right)^2+4$
$\displaystyle =5^2+4$
$\displaystyle =29$
$\displaystyle \therefore a+\frac{1}{a}=\pm\sqrt{29}$

$\displaystyle \text{(iii) }a^2-\frac{1}{a^2}=\left(a-\frac{1}{a}\right)\left(a+\frac{1}{a}\right)$
$\displaystyle =5(\pm\sqrt{29})$
$\displaystyle =\pm5\sqrt{29}$
$\displaystyle \therefore a^2-\frac{1}{a^2}=\pm5\sqrt{29}$
$\displaystyle \\$

$\displaystyle \textbf{Question 14: }\text{If }3x+4y=16\text{ and }xy=4,\text{ find the value of}$
$\displaystyle 9x^2+16y^2.$
$\displaystyle \text{Answer:}$
$\displaystyle (3x+4y)^2=9x^2+16y^2+24xy$
$\displaystyle 16^2=9x^2+16y^2+24(4)$
$\displaystyle 256=9x^2+16y^2+96$
$\displaystyle 9x^2+16y^2=256-96$
$\displaystyle =160$
$\displaystyle \therefore \text{The required value is }160.$
$\displaystyle \\$

$\displaystyle \textbf{Question 15: }\text{The number }x\text{ is }2\text{ more than the number }y.\text{ If}$
$\displaystyle \text{the sum of the squares of }x\text{ and }y\text{ is }34,\text{ find the product of }x\text{ and }y.$
$\displaystyle \text{Answer:}$
$\displaystyle x-y=2$
$\displaystyle x^2+y^2=34$
$\displaystyle (x-y)^2=x^2+y^2-2xy$
$\displaystyle 2^2=34-2xy$
$\displaystyle 4=34-2xy$
$\displaystyle 2xy=30$
$\displaystyle xy=15$
$\displaystyle \therefore \text{The product of }x\text{ and }y\text{ is }15.$
$\displaystyle \\$

$\displaystyle \textbf{Question 16: }\text{The difference between two positive numbers is }5$
$\displaystyle \text{and the sum of their squares is }73.\text{ Find the product of these numbers.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the two positive numbers be }x\text{ and }y,\text{ where }x>y.$
$\displaystyle x-y=5$
$\displaystyle x^2+y^2=73$
$\displaystyle (x-y)^2=x^2+y^2-2xy$
$\displaystyle 5^2=73-2xy$
$\displaystyle 25=73-2xy$
$\displaystyle 2xy=73-25$
$\displaystyle =48$
$\displaystyle xy=24$
$\displaystyle \therefore \text{The product of the two numbers is }24.$
$\displaystyle \\$

$\displaystyle \textbf{Exercise 4(B)}$

$\displaystyle \textbf{Question 1: }\text{Find the cube of:}$
$\displaystyle \text{(i) }3a-2b \qquad \text{(ii) }5a+3b \qquad \text{(iii) }2a+\frac{1}{2a}\ (a\ne0) \qquad \text{(iv) }3a-\frac{1}{a}\ (a\ne0)$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }\ (3a-2b)^3$
$\displaystyle =(3a)^3-3(3a)^2(2b)+3(3a)(2b)^2-(2b)^3$
$\displaystyle =27a^3-54a^2b+36ab^2-8b^3$
$\displaystyle \therefore (3a-2b)^3=27a^3-54a^2b+36ab^2-8b^3.$

$\displaystyle \text{(ii) }\ (5a+3b)^3$
$\displaystyle =(5a)^3+3(5a)^2(3b)+3(5a)(3b)^2+(3b)^3$
$\displaystyle =125a^3+225a^2b+135ab^2+27b^3$
$\displaystyle \therefore (5a+3b)^3=125a^3+225a^2b+135ab^2+27b^3.$

$\displaystyle \text{(iii) }\ \left(2a+\frac{1}{2a}\right)^3$
$\displaystyle =(2a)^3+3(2a)^2\left(\frac{1}{2a}\right)+3(2a)\left(\frac{1}{2a}\right)^2+\left(\frac{1}{2a}\right)^3$
$\displaystyle =8a^3+6a+\frac{3}{2a}+\frac{1}{8a^3}$
$\displaystyle \therefore \left(2a+\frac{1}{2a}\right)^3=8a^3+6a+\frac{3}{2a}+\frac{1}{8a^3}.$

$\displaystyle \text{(iv) }\ \left(3a-\frac{1}{a}\right)^3$
$\displaystyle =(3a)^3-3(3a)^2\left(\frac{1}{a}\right)+3(3a)\left(\frac{1}{a}\right)^2-\left(\frac{1}{a}\right)^3$
$\displaystyle =27a^3-27a+\frac{9}{a}-\frac{1}{a^3}$
$\displaystyle \therefore \left(3a-\frac{1}{a}\right)^3=27a^3-27a+\frac{9}{a}-\frac{1}{a^3}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{If }a^2+\frac{1}{a^2}=47\text{ and }a\ne0,\text{ find:}$
$\displaystyle \text{(i) }a+\frac{1}{a}\qquad \text{(ii) }a^3+\frac{1}{a^3}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }\left(a+\frac{1}{a}\right)^2=a^2+\frac{1}{a^2}+2$
$\displaystyle =47+2$
$\displaystyle =49$
$\displaystyle \therefore a+\frac{1}{a}=\pm7$

$\displaystyle \text{(ii) }a^3+\frac{1}{a^3}=\left(a+\frac{1}{a}\right)^3-3\left(a+\frac{1}{a}\right)$
$\displaystyle =(\pm7)^3-3(\pm7)$
$\displaystyle =\pm343\mp21$
$\displaystyle =\pm322$
$\displaystyle \therefore a^3+\frac{1}{a^3}=\pm322.$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{If }a^2+\frac{1}{a^2}=18\text{ and }a\ne0,\text{ find:}$
$\displaystyle \text{(i) }a-\frac{1}{a}\qquad \text{(ii) }a^3-\frac{1}{a^3}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }\left(a-\frac{1}{a}\right)^2=a^2+\frac{1}{a^2}-2$
$\displaystyle =18-2$
$\displaystyle =16$
$\displaystyle \therefore a-\frac{1}{a}=\pm4$

$\displaystyle \text{(ii) }a^3-\frac{1}{a^3}=\left(a-\frac{1}{a}\right)^3+3\left(a-\frac{1}{a}\right)$
$\displaystyle =(\pm4)^3+3(\pm4)$
$\displaystyle =\pm64\pm12$
$\displaystyle =\pm76$
$\displaystyle \therefore a^3-\frac{1}{a^3}=\pm76.$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{If }a+\frac{1}{a}=p\text{ and }a\ne0;\text{ then show that:}$
$\displaystyle a^3+\frac{1}{a^3}=p(p^2-3)$
$\displaystyle \text{Answer:}$
$\displaystyle a+\frac{1}{a}=p$
$\displaystyle \text{Using the identity,}$
$\displaystyle a^3+\frac{1}{a^3}=\left(a+\frac{1}{a}\right)^3-3\left(a+\frac{1}{a}\right)$
$\displaystyle =p^3-3p$
$\displaystyle =p(p^2-3)$
$\displaystyle \therefore a^3+\frac{1}{a^3}=p(p^2-3).$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{If }a+2b=5;\text{ then show that:}$
$\displaystyle a^3+8b^3+30ab=125.$
$\displaystyle \text{Answer:}$
$\displaystyle a+2b=5$
$\displaystyle \text{Cubing both sides,}$
$\displaystyle (a+2b)^3=5^3$
$\displaystyle a^3+3(a)^2(2b)+3(a)(2b)^2+(2b)^3=125$
$\displaystyle a^3+6a^2b+12ab^2+8b^3=125$
$\displaystyle a^3+8b^3+6ab(a+2b)=125$
$\displaystyle a^3+8b^3+6ab(5)=125$
$\displaystyle a^3+8b^3+30ab=125$
$\displaystyle \therefore a^3+8b^3+30ab=125.$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\text{If }\left(a+\frac{1}{a}\right)^2=3\text{ and }a\ne0;\text{ then show that:}$
$\displaystyle a^3+\frac{1}{a^3}=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \left(a+\frac{1}{a}\right)^2=3$
$\displaystyle \therefore a+\frac{1}{a}=\pm\sqrt3$
$\displaystyle \text{Using the identity,}$
$\displaystyle a^3+\frac{1}{a^3}=\left(a+\frac{1}{a}\right)^3-3\left(a+\frac{1}{a}\right)$
$\displaystyle =\left(a+\frac{1}{a}\right)\left[\left(a+\frac{1}{a}\right)^2-3\right]$
$\displaystyle =\left(a+\frac{1}{a}\right)(3-3)$
$\displaystyle =0$
$\displaystyle \therefore a^3+\frac{1}{a^3}=0.$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: }\text{If }a+2b+c=0;\text{ then show that:}$
$\displaystyle a^3+8b^3+c^3=6abc.$
$\displaystyle \text{Answer:}$
$\displaystyle a+2b+c=0$
$\displaystyle \text{Using the identity, if }x+y+z=0,\text{ then }x^3+y^3+z^3=3xyz.$
$\displaystyle \text{Taking }x=a,\ y=2b\text{ and }z=c,\text{ we get}$
$\displaystyle a^3+(2b)^3+c^3=3(a)(2b)(c)$
$\displaystyle a^3+8b^3+c^3=6abc$
$\displaystyle \therefore a^3+8b^3+c^3=6abc.$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: }\text{Use property to evaluate:}$
$\displaystyle \text{(i) }13^3+(-8)^3+(-5)^3 \qquad \text{(ii) }7^3+3^3+(-10)^3$
$\displaystyle \text{(iii) }9^3-5^3-4^3 \qquad \text{(iv) }38^3+(-26)^3+(-12)^3$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) Since }13+(-8)+(-5)=0,\text{ using }a^3+b^3+c^3=3abc,$
$\displaystyle 13^3+(-8)^3+(-5)^3=3(13)(-8)(-5)$
$\displaystyle =1,560$
$\displaystyle \therefore 13^3+(-8)^3+(-5)^3=1,560.$

$\displaystyle \text{(ii) Since }7+3+(-10)=0,\text{ using }a^3+b^3+c^3=3abc,$
$\displaystyle 7^3+3^3+(-10)^3=3(7)(3)(-10)$
$\displaystyle =-630$
$\displaystyle \therefore 7^3+3^3+(-10)^3=-630.$

$\displaystyle \text{(iii) Since }9+(-5)+(-4)=0,\text{ using }a^3+b^3+c^3=3abc,$
$\displaystyle 9^3-5^3-4^3=3(9)(-5)(-4)$
$\displaystyle =540$
$\displaystyle \therefore 9^3-5^3-4^3=540.$

$\displaystyle \text{(iv) Since }38+(-26)+(-12)=0,\text{ using }a^3+b^3+c^3=3abc,$
$\displaystyle 38^3+(-26)^3+(-12)^3=3(38)(-26)(-12)$
$\displaystyle =35,568$
$\displaystyle \therefore 38^3+(-26)^3+(-12)^3=35,568.$
$\displaystyle \\$

$\displaystyle \textbf{Question 9: }\text{If }a\ne0\text{ and }a-\frac{1}{a}=3;\text{ find:}$
$\displaystyle \text{(i) }a^2+\frac{1}{a^2}\qquad \text{(ii) }a^3-\frac{1}{a^3}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }\left(a-\frac{1}{a}\right)^2=a^2+\frac{1}{a^2}-2$
$\displaystyle 3^2=a^2+\frac{1}{a^2}-2$
$\displaystyle a^2+\frac{1}{a^2}=9+2$
$\displaystyle =11$
$\displaystyle \therefore a^2+\frac{1}{a^2}=11.$

$\displaystyle \text{(ii) }a^3-\frac{1}{a^3}=\left(a-\frac{1}{a}\right)^3+3\left(a-\frac{1}{a}\right)$
$\displaystyle =3^3+3(3)$
$\displaystyle =27+9$
$\displaystyle =36$
$\displaystyle \therefore a^3-\frac{1}{a^3}=36.$
$\displaystyle \\$

$\displaystyle \textbf{Question 10: }\text{If }a\ne0\text{ and }a-\frac{1}{a}=4;\text{ find:}$
$\displaystyle \text{(i) }a^2+\frac{1}{a^2}\qquad \text{(ii) }a^4+\frac{1}{a^4}\qquad \text{(iii) }a^3-\frac{1}{a^3}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }\left(a-\frac{1}{a}\right)^2=a^2+\frac{1}{a^2}-2$
$\displaystyle 4^2=a^2+\frac{1}{a^2}-2$
$\displaystyle a^2+\frac{1}{a^2}=16+2$
$\displaystyle =18$
$\displaystyle \therefore a^2+\frac{1}{a^2}=18.$

$\displaystyle \text{(ii) }\left(a^2+\frac{1}{a^2}\right)^2=a^4+\frac{1}{a^4}+2$
$\displaystyle 18^2=a^4+\frac{1}{a^4}+2$
$\displaystyle a^4+\frac{1}{a^4}=324-2$
$\displaystyle =322$
$\displaystyle \therefore a^4+\frac{1}{a^4}=322.$

$\displaystyle \text{(iii) }a^3-\frac{1}{a^3}=\left(a-\frac{1}{a}\right)^3+3\left(a-\frac{1}{a}\right)$
$\displaystyle =4^3+3(4)$
$\displaystyle =64+12$
$\displaystyle =76$
$\displaystyle \therefore a^3-\frac{1}{a^3}=76.$
$\displaystyle \\$

$\displaystyle \textbf{Question 11: }\text{If }x\ne0\text{ and }x+\frac{1}{x}=2;\text{ then show that:}$
$\displaystyle x^2+\frac{1}{x^2}=x^3+\frac{1}{x^3}=x^4+\frac{1}{x^4}$
$\displaystyle \text{Answer:}$
$\displaystyle x+\frac{1}{x}=2$
$\displaystyle x^2+\frac{1}{x^2}=\left(x+\frac{1}{x}\right)^2-2$
$\displaystyle =2^2-2$
$\displaystyle =2\qquad\ldots(1)$
$\displaystyle x^3+\frac{1}{x^3}=\left(x+\frac{1}{x}\right)^3-3\left(x+\frac{1}{x}\right)$
$\displaystyle =2^3-3(2)$
$\displaystyle =8-6$
$\displaystyle =2\qquad\ldots(2)$
$\displaystyle x^4+\frac{1}{x^4}=\left(x^2+\frac{1}{x^2}\right)^2-2$
$\displaystyle =2^2-2$
$\displaystyle =2\qquad\ldots(3)$
$\displaystyle \text{From (1), (2) and (3),}$
$\displaystyle \therefore x^2+\frac{1}{x^2}=x^3+\frac{1}{x^3}=x^4+\frac{1}{x^4}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 12: }\text{If }2x-3y=10\text{ and }xy=16,\text{ find the value of}$
$\displaystyle 8x^3-27y^3.$
$\displaystyle \text{Answer:}$
$\displaystyle 8x^3-27y^3=(2x)^3-(3y)^3$
$\displaystyle \text{Using }a^3-b^3=(a-b)^3+3ab(a-b),$
$\displaystyle (2x)^3-(3y)^3=(2x-3y)^3+3(2x)(3y)(2x-3y)$
$\displaystyle =10^3+3(6xy)(10)$
$\displaystyle =1,000+180xy$
$\displaystyle =1,000+180(16)$
$\displaystyle =1,000+2,880$
$\displaystyle =3,880$
$\displaystyle \therefore \text{The required value is }3,880.$
$\displaystyle \\$

$\displaystyle \textbf{Question 13: }\text{Expand:}$
$\displaystyle \text{(i) }(3x+5y+2z)(3x-5y+2z)$
$\displaystyle \text{(ii) }(3x-5y-2z)(3x-5y+2z)$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }(3x+5y+2z)(3x-5y+2z)$
$\displaystyle =\big[(3x+2z)+5y\big]\big[(3x+2z)-5y\big]$
$\displaystyle =(3x+2z)^2-(5y)^2$
$\displaystyle =9x^2+12xz+4z^2-25y^2$
$\displaystyle \therefore (3x+5y+2z)(3x-5y+2z)=9x^2-25y^2+12xz+4z^2.$

$\displaystyle \text{(ii) }(3x-5y-2z)(3x-5y+2z)$
$\displaystyle =\big[(3x-5y)-2z\big]\big[(3x-5y)+2z\big]$
$\displaystyle =(3x-5y)^2-(2z)^2$
$\displaystyle =9x^2-30xy+25y^2-4z^2$
$\displaystyle \therefore (3x-5y-2z)(3x-5y+2z)=9x^2-30xy+25y^2-4z^2.$
$\displaystyle \\$

$\displaystyle \textbf{Question 14: }\text{The sum of two numbers is }9\text{ and their product is }20.\text{ Find the sum of their:}$
$\displaystyle \text{(i) squares}\qquad \text{(ii) cubes}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the two numbers be }a\text{ and }b.$
$\displaystyle a+b=9,\qquad ab=20$
$\displaystyle \text{(i) }a^2+b^2=(a+b)^2-2ab$
$\displaystyle =9^2-2(20)$
$\displaystyle =81-40$
$\displaystyle =41$
$\displaystyle \therefore \text{The sum of the squares is }41.$

$\displaystyle \text{(ii) }a^3+b^3=(a+b)^3-3ab(a+b)$
$\displaystyle =9^3-3(20)(9)$
$\displaystyle =729-540$
$\displaystyle =189$
$\displaystyle \therefore \text{The sum of the cubes is }189.$
$\displaystyle \\$

$\displaystyle \textbf{Question 15: }\text{Two positive numbers }x\text{ and }y\text{ are such that }x>y.$
$\displaystyle \text{If the difference of these numbers is }5\text{ and their product is }24,\text{ find:}$
$\displaystyle \text{(i) sum of these numbers}\qquad \text{(ii) difference of their cubes}\qquad \text{(iii) sum of their cubes}$
$\displaystyle \text{Answer:}$
$\displaystyle x-y=5,\qquad xy=24$
$\displaystyle \text{(i) }(x+y)^2=(x-y)^2+4xy$
$\displaystyle =5^2+4(24)$
$\displaystyle =25+96$
$\displaystyle =121$
$\displaystyle \therefore x+y=11.$

$\displaystyle \text{(ii) }x^3-y^3=(x-y)^3+3xy(x-y)$
$\displaystyle =5^3+3(24)(5)$
$\displaystyle =125+360$
$\displaystyle =485$
$\displaystyle \therefore \text{The difference of their cubes is }485.$

$\displaystyle \text{(iii) }x^3+y^3=(x+y)^3-3xy(x+y)$
$\displaystyle =11^3-3(24)(11)$
$\displaystyle =1331-792$
$\displaystyle =539$
$\displaystyle \therefore \text{The sum of their cubes is }539.$
$\displaystyle \\$

$\displaystyle \textbf{Exercise 4(C)}$

$\displaystyle \textbf{Question 1: }\text{Expand:}$
$\displaystyle \text{(i) }(x+8)(x+10)\qquad \text{(ii) }(x+8)(x-10)\qquad \text{(iii) }(x-8)(x+10)\qquad \text{(iv) }(x-8)(x-10)$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }(x+8)(x+10)$
$\displaystyle =x^2+(8+10)x+(8)(10)$
$\displaystyle =x^2+18x+80$
$\displaystyle \therefore (x+8)(x+10)=x^2+18x+80.$

$\displaystyle \text{(ii) }(x+8)(x-10)$
$\displaystyle =x^2+(8-10)x-(8)(10)$
$\displaystyle =x^2-2x-80$
$\displaystyle \therefore (x+8)(x-10)=x^2-2x-80.$

$\displaystyle \text{(iii) }(x-8)(x+10)$
$\displaystyle =x^2+(10-8)x-(8)(10)$
$\displaystyle =x^2+2x-80$
$\displaystyle \therefore (x-8)(x+10)=x^2+2x-80.$

$\displaystyle \text{(iv) }(x-8)(x-10)$
$\displaystyle =x^2-(8+10)x+(8)(10)$
$\displaystyle =x^2-18x+80$
$\displaystyle \therefore (x-8)(x-10)=x^2-18x+80.$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{Expand:}$
$\displaystyle \text{(i) }\left(2x-\frac{1}{x}\right)\left(3x+\frac{2}{x}\right)\qquad \text{(ii) }\left(3a+\frac{2}{b}\right)\left(2a-\frac{3}{b}\right)$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }\left(2x-\frac{1}{x}\right)\left(3x+\frac{2}{x}\right)$
$\displaystyle =6x^2+\frac{4x}{x}-\frac{3x}{x}-\frac{2}{x^2}$
$\displaystyle =6x^2+1-\frac{2}{x^2}$
$\displaystyle \therefore \left(2x-\frac{1}{x}\right)\left(3x+\frac{2}{x}\right)=6x^2+1-\frac{2}{x^2}.$

$\displaystyle \text{(ii) }\left(3a+\frac{2}{b}\right)\left(2a-\frac{3}{b}\right)$
$\displaystyle =6a^2-\frac{9a}{b}+\frac{4a}{b}-\frac{6}{b^2}$
$\displaystyle =6a^2-\frac{5a}{b}-\frac{6}{b^2}$
$\displaystyle \therefore \left(3a+\frac{2}{b}\right)\left(2a-\frac{3}{b}\right)=6a^2-\frac{5a}{b}-\frac{6}{b^2}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{Expand:}$
$\displaystyle \text{(i) }(x+y-z)^2\qquad \text{(ii) }(x-2y+2)^2\qquad \text{(iii) }(5a-3b+c)^2$
$\displaystyle \text{(iv) }(5x-3y-2)^2\qquad \text{(v) }\left(x-\frac{1}{x}+5\right)^2$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }(x+y-z)^2$
$\displaystyle =x^2+y^2+z^2+2xy-2yz-2xz$
$\displaystyle \therefore (x+y-z)^2=x^2+y^2+z^2+2xy-2yz-2xz.$

$\displaystyle \text{(ii) }(x-2y+2)^2$
$\displaystyle =x^2+(-2y)^2+2^2+2(x)(-2y)+2(-2y)(2)+2(x)(2)$
$\displaystyle =x^2+4y^2+4-4xy-8y+4x$
$\displaystyle \therefore (x-2y+2)^2=x^2+4y^2+4-4xy+4x-8y.$

$\displaystyle \text{(iii) }(5a-3b+c)^2$
$\displaystyle =(5a)^2+(-3b)^2+c^2+2(5a)(-3b)+2(-3b)(c)+2(5a)(c)$
$\displaystyle =25a^2+9b^2+c^2-30ab-6bc+10ac$
$\displaystyle \therefore (5a-3b+c)^2=25a^2+9b^2+c^2-30ab-6bc+10ac.$

$\displaystyle \text{(iv) }(5x-3y-2)^2$
$\displaystyle =(5x)^2+(-3y)^2+(-2)^2+2(5x)(-3y)+2(-3y)(-2)+2(5x)(-2)$
$\displaystyle =25x^2+9y^2+4-30xy+12y-20x$
$\displaystyle \therefore (5x-3y-2)^2=25x^2+9y^2+4-30xy-20x+12y.$

$\displaystyle \text{(v) }\left(x-\frac{1}{x}+5\right)^2$
$\displaystyle =x^2+\frac{1}{x^2}+25+2(x)\left(-\frac{1}{x}\right)+2\left(-\frac{1}{x}\right)(5)+2(x)(5)$
$\displaystyle =x^2+\frac{1}{x^2}+23+10x-\frac{10}{x}$
$\displaystyle \therefore \left(x-\frac{1}{x}+5\right)^2=x^2+\frac{1}{x^2}+23+10x-\frac{10}{x}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{If }a+b+c=12\text{ and }a^2+b^2+c^2=50,\text{ find}$
$\displaystyle ab+bc+ca.$
$\displaystyle \text{Answer:}$
$\displaystyle (a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$
$\displaystyle 12^2=50+2(ab+bc+ca)$
$\displaystyle 144=50+2(ab+bc+ca)$
$\displaystyle 2(ab+bc+ca)=144-50$
$\displaystyle =94$
$\displaystyle ab+bc+ca=47$
$\displaystyle \therefore \text{The required value is }47.$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{If }a^2+b^2+c^2=35\text{ and }ab+bc+ca=23,\text{ find}$
$\displaystyle a+b+c.$
$\displaystyle \text{Answer:}$
$\displaystyle (a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$
$\displaystyle =35+2(23)$
$\displaystyle =35+46$
$\displaystyle =81$
$\displaystyle \therefore a+b+c=\pm9$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\text{If }a+b+c=p\text{ and }ab+bc+ca=q,\text{ find}$
$\displaystyle a^2+b^2+c^2.$
$\displaystyle \text{Answer:}$
$\displaystyle (a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$
$\displaystyle p^2=a^2+b^2+c^2+2q$
$\displaystyle a^2+b^2+c^2=p^2-2q$
$\displaystyle \therefore \text{The required value is }p^2-2q.$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: }\text{If }a^2+b^2+c^2=50\text{ and }ab+bc+ca=47,$
$\displaystyle \text{find }a+b+c.$
$\displaystyle \text{Answer:}$
$\displaystyle (a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$
$\displaystyle =50+2(47)$
$\displaystyle =50+94$
$\displaystyle =144$
$\displaystyle \therefore a+b+c=\pm12$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: }\text{If }x+y-z=4\text{ and }x^2+y^2+z^2=30,\text{ then find the}$
$\displaystyle \text{value of }xy-yz-zx.$
$\displaystyle \text{Answer:}$
$\displaystyle (x+y-z)^2=x^2+y^2+z^2+2xy-2yz-2zx$
$\displaystyle =x^2+y^2+z^2+2(xy-yz-zx)$
$\displaystyle 4^2=30+2(xy-yz-zx)$
$\displaystyle 16=30+2(xy-yz-zx)$
$\displaystyle 2(xy-yz-zx)=16-30$
$\displaystyle =-14$
$\displaystyle xy-yz-zx=-7$
$\displaystyle \therefore \text{The required value is }-7.$
$\displaystyle \\$

$\displaystyle \textbf{Exercise 4(D)}$

$\displaystyle \textbf{Question 1: }\text{If }x+2y+3z=0\text{ and}$
$\displaystyle x^3+4y^3+9z^3=18xyz,\text{ evaluate:}$
$\displaystyle \frac{(x+2y)^2}{xy}+\frac{(2y+3z)^2}{yz}+\frac{(3z+x)^2}{zx}$
$\displaystyle \text{Answer:}$
$\displaystyle x+2y+3z=0$
$\displaystyle \therefore x+2y=-3z,\qquad 2y+3z=-x,\qquad 3z+x=-2y$
$\displaystyle \frac{(x+2y)^2}{xy}+\frac{(2y+3z)^2}{yz}+\frac{(3z+x)^2}{zx}$
$\displaystyle =\frac{(-3z)^2}{xy}+\frac{(-x)^2}{yz}+\frac{(-2y)^2}{zx}$
$\displaystyle =\frac{9z^2}{xy}+\frac{x^2}{yz}+\frac{4y^2}{zx}$
$\displaystyle =\frac{9z^3+x^3+4y^3}{xyz}$
$\displaystyle =\frac{x^3+4y^3+9z^3}{xyz}$
$\displaystyle =\frac{18xyz}{xyz}$
$\displaystyle =18$
$\displaystyle \therefore \text{The required value is }18.$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{If }a+\frac{1}{a}=m\text{ and }a\ne0,\text{ find in terms of }m,\text{ the value of:}$
$\displaystyle \text{(i) }a-\frac{1}{a}\qquad \text{(ii) }a^2-\frac{1}{a^2}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }\left(a+\frac{1}{a}\right)^2-\left(a-\frac{1}{a}\right)^2=4$
$\displaystyle m^2-\left(a-\frac{1}{a}\right)^2=4$
$\displaystyle \left(a-\frac{1}{a}\right)^2=m^2-4$
$\displaystyle \therefore a-\frac{1}{a}=\pm\sqrt{m^2-4}$

$\displaystyle \text{(ii) }a^2-\frac{1}{a^2}=\left(a+\frac{1}{a}\right)\left(a-\frac{1}{a}\right)$
$\displaystyle =m\left(\pm\sqrt{m^2-4}\right)$
$\displaystyle =\pm m\sqrt{m^2-4}$
$\displaystyle \therefore a^2-\frac{1}{a^2}=\pm m\sqrt{m^2-4}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{In the expansion of }(2x^2-8)(x-4)^2,\text{ find the value of:}$
$\displaystyle \text{(i) coefficient of }x^3\qquad \text{(ii) coefficient of }x^2\qquad \text{(iii) constant term.}$
$\displaystyle \text{Answer:}$
$\displaystyle (2x^2-8)(x-4)^2$
$\displaystyle =(2x^2-8)(x^2-8x+16)$
$\displaystyle =2x^2(x^2-8x+16)-8(x^2-8x+16)$
$\displaystyle =2x^4-16x^3+32x^2-8x^2+64x-128$
$\displaystyle =2x^4-16x^3+24x^2+64x-128$
$\displaystyle \text{(i) Coefficient of }x^3=-16$

$\displaystyle \text{(ii) Coefficient of }x^2=24$

$\displaystyle \text{(iii) Constant term}=-128$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{If }x>0\text{ and }x^2+\frac{1}{9x^2}=\frac{25}{36},\text{ find: }x^3+\frac{1}{27x^3}.$
$\displaystyle \text{Answer:}$
$\displaystyle x^2+\frac{1}{9x^2}=\frac{25}{36}$
$\displaystyle \left(x+\frac{1}{3x}\right)^2=x^2+\frac{1}{9x^2}+\frac{2}{3}$
$\displaystyle =\frac{25}{36}+\frac{24}{36}$
$\displaystyle =\frac{49}{36}$
$\displaystyle \text{Since }x>0,\quad x+\frac{1}{3x}=\frac{7}{6}$
$\displaystyle x^3+\frac{1}{27x^3}=\left(x+\frac{1}{3x}\right)^3-3\left(x\right)\left(\frac{1}{3x}\right)\left(x+\frac{1}{3x}\right)$
$\displaystyle =\left(\frac{7}{6}\right)^3-\left(\frac{7}{6}\right)$
$\displaystyle =\frac{343}{216}-\frac{252}{216}$
$\displaystyle =\frac{91}{216}$
$\displaystyle \therefore x^3+\frac{1}{27x^3}=\frac{91}{216}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{If }2(x^2+1)=5x,\text{ find:}$
$\displaystyle \text{(i) }x-\frac{1}{x}\qquad \text{(ii) }x^3-\frac{1}{x^3}$
$\displaystyle \text{Answer:}$
$\displaystyle 2(x^2+1)=5x$
$\displaystyle \text{Dividing each term by }2x,\text{ we get}$
$\displaystyle x+\frac{1}{x}=\frac{5}{2}$
$\displaystyle \text{(i) }\left(x-\frac{1}{x}\right)^2=\left(x+\frac{1}{x}\right)^2-4$
$\displaystyle =\left(\frac{5}{2}\right)^2-4$
$\displaystyle =\frac{25}{4}-\frac{16}{4}$
$\displaystyle =\frac{9}{4}$
$\displaystyle \therefore x-\frac{1}{x}=\pm\frac{3}{2}$

$\displaystyle \text{(ii) }x^3-\frac{1}{x^3}=\left(x-\frac{1}{x}\right)^3+3\left(x-\frac{1}{x}\right)$
$\displaystyle =\left(\pm\frac{3}{2}\right)^3+3\left(\pm\frac{3}{2}\right)$
$\displaystyle =\pm\frac{27}{8}\pm\frac{36}{8}$
$\displaystyle =\pm\frac{63}{8}$
$\displaystyle \therefore x^3-\frac{1}{x^3}=\pm\frac{63}{8}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\text{If }a^2+b^2=34\text{ and }ab=12,\text{ find:}$
$\displaystyle \text{(i) }3(a+b)^2+5(a-b)^2\qquad \text{(ii) }7(a-b)^2-2(a+b)^2$
$\displaystyle \text{Answer:}$
$\displaystyle (a+b)^2=a^2+b^2+2ab$
$\displaystyle =34+2(12)$
$\displaystyle =58$
$\displaystyle (a-b)^2=a^2+b^2-2ab$
$\displaystyle =34-2(12)$
$\displaystyle =10$
$\displaystyle \text{(i) }3(a+b)^2+5(a-b)^2$
$\displaystyle =3(58)+5(10)$
$\displaystyle =174+50$
$\displaystyle =224$
$\displaystyle \therefore \text{The required value is }224.$

$\displaystyle \text{(ii) }7(a-b)^2-2(a+b)^2$
$\displaystyle =7(10)-2(58)$
$\displaystyle =70-116$
$\displaystyle =-46$
$\displaystyle \therefore \text{The required value is }-46.$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: }\text{If }3x-\frac{4}{x}=4\text{ and }x\ne0,\text{ find: }27x^3-\frac{64}{x^3}.$
$\displaystyle \text{Answer:}$
$\displaystyle 27x^3-\frac{64}{x^3}=(3x)^3-\left(\frac{4}{x}\right)^3$
$\displaystyle \text{Using }a^3-b^3=(a-b)^3+3ab(a-b),$
$\displaystyle (3x)^3-\left(\frac{4}{x}\right)^3$
$\displaystyle =\left(3x-\frac{4}{x}\right)^3+3(3x)\left(\frac{4}{x}\right)\left(3x-\frac{4}{x}\right)$
$\displaystyle =4^3+3(12)(4)$
$\displaystyle =64+144$
$\displaystyle =208$
$\displaystyle \therefore 27x^3-\frac{64}{x^3}=208.$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: }\text{If }x^2+\frac{1}{x^2}=7\text{ and }x\ne0,\text{ find the value of:}$
$\displaystyle 7x^3+8x-\frac{7}{x^3}-\frac{8}{x}.$
$\displaystyle \text{Answer:}$
$\displaystyle \left(x-\frac{1}{x}\right)^2=x^2+\frac{1}{x^2}-2$
$\displaystyle =7-2$
$\displaystyle =5$
$\displaystyle \therefore x-\frac{1}{x}=\pm\sqrt5$
$\displaystyle x^3-\frac{1}{x^3}=\left(x-\frac{1}{x}\right)^3+3\left(x-\frac{1}{x}\right)$
$\displaystyle =\left(x-\frac{1}{x}\right)\left[\left(x-\frac{1}{x}\right)^2+3\right]$
$\displaystyle =\left(x-\frac{1}{x}\right)(5+3)$
$\displaystyle =8\left(x-\frac{1}{x}\right)$
$\displaystyle 7x^3+8x-\frac{7}{x^3}-\frac{8}{x}$
$\displaystyle =7\left(x^3-\frac{1}{x^3}\right)+8\left(x-\frac{1}{x}\right)$
$\displaystyle =7\left[8\left(x-\frac{1}{x}\right)\right]+8\left(x-\frac{1}{x}\right)$
$\displaystyle =64\left(x-\frac{1}{x}\right)$
$\displaystyle =\pm64\sqrt5$
$\displaystyle \therefore \text{The required value is }\pm64\sqrt5.$
$\displaystyle \\$

$\displaystyle \textbf{Question 9: }\text{If }x=\frac{1}{x-5}\text{ and }x\ne5,\text{ find: }x^2-\frac{1}{x^2}.$
$\displaystyle \text{Answer:}$
$\displaystyle x=\frac{1}{x-5}$
$\displaystyle x(x-5)=1$
$\displaystyle x^2-5x=1$
$\displaystyle \text{Dividing each term by }x,\text{ we get}$
$\displaystyle x-5=\frac{1}{x}$
$\displaystyle \therefore x-\frac{1}{x}=5$
$\displaystyle \left(x+\frac{1}{x}\right)^2=\left(x-\frac{1}{x}\right)^2+4$
$\displaystyle =5^2+4$
$\displaystyle =29$
$\displaystyle \therefore x+\frac{1}{x}=\pm\sqrt{29}$
$\displaystyle x^2-\frac{1}{x^2}=\left(x-\frac{1}{x}\right)\left(x+\frac{1}{x}\right)$
$\displaystyle =5(\pm\sqrt{29})$
$\displaystyle =\pm5\sqrt{29}$
$\displaystyle \therefore x^2-\frac{1}{x^2}=\pm5\sqrt{29}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 10: }\text{If }x=\frac{1}{5-x}\text{ and }x\ne5,\text{ find: }x^3+\frac{1}{x^3}.$
$\displaystyle \text{Answer:}$
$\displaystyle x=\frac{1}{5-x}$
$\displaystyle x(5-x)=1$
$\displaystyle 5x-x^2=1$
$\displaystyle x^2-5x+1=0$
$\displaystyle \text{Dividing each term by }x,\text{ we get}$
$\displaystyle x-5+\frac{1}{x}=0$
$\displaystyle \therefore x+\frac{1}{x}=5$
$\displaystyle x^3+\frac{1}{x^3}=\left(x+\frac{1}{x}\right)^3-3\left(x+\frac{1}{x}\right)$
$\displaystyle =5^3-3(5)$
$\displaystyle =125-15$
$\displaystyle =110$
$\displaystyle \therefore x^3+\frac{1}{x^3}=110.$
$\displaystyle \\$

$\displaystyle \textbf{Question 11: }\text{If }3a+5b+4c=0,\text{ show that:}$
$\displaystyle 27a^3+125b^3+64c^3=180abc.$
$\displaystyle \text{Answer:}$
$\displaystyle 3a+5b+4c=0$
$\displaystyle \text{Using the identity, if }x+y+z=0,\text{ then }x^3+y^3+z^3=3xyz.$
$\displaystyle \text{Taking }x=3a,\ y=5b\text{ and }z=4c,\text{ we get}$
$\displaystyle (3a)^3+(5b)^3+(4c)^3=3(3a)(5b)(4c)$
$\displaystyle 27a^3+125b^3+64c^3=180abc$
$\displaystyle \therefore 27a^3+125b^3+64c^3=180abc.$
$\displaystyle \\$

$\displaystyle \textbf{Question 12: }\text{The sum of two numbers is }7\text{ and the sum of their}$
$\displaystyle \text{cubes is }133.\text{ Find the sum of their squares.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the two numbers be }a\text{ and }b.$
$\displaystyle a+b=7,\qquad a^3+b^3=133$
$\displaystyle a^3+b^3=(a+b)^3-3ab(a+b)$
$\displaystyle 133=7^3-3ab(7)$
$\displaystyle 133=343-21ab$
$\displaystyle 21ab=343-133$
$\displaystyle =210$
$\displaystyle ab=10$
$\displaystyle a^2+b^2=(a+b)^2-2ab$
$\displaystyle =7^2-2(10)$
$\displaystyle =49-20$
$\displaystyle =29$
$\displaystyle \therefore \text{The sum of their squares is }29.$
$\displaystyle \\$

$\displaystyle \textbf{Question 13: }\text{In each of the following, find the value of }a:$
$\displaystyle \text{(i) }4x^2+ax+9=(2x+3)^2$
$\displaystyle \text{(ii) }4x^2+ax+9=(2x-3)^2$
$\displaystyle \text{(iii) }9x^2+(7a-5)x+25=(3x+5)^2$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }4x^2+ax+9=(2x+3)^2$
$\displaystyle =4x^2+12x+9$
$\displaystyle \therefore a=12$

$\displaystyle \text{(ii) }4x^2+ax+9=(2x-3)^2$
$\displaystyle =4x^2-12x+9$
$\displaystyle \therefore a=-12$

$\displaystyle \text{(iii) }9x^2+(7a-5)x+25=(3x+5)^2$
$\displaystyle =9x^2+30x+25$
$\displaystyle \therefore 7a-5=30$
$\displaystyle 7a=35$
$\displaystyle a=5$
$\displaystyle \\$

$\displaystyle \textbf{Question 14: }\text{If }\frac{x^2+1}{x}=3\frac{1}{3}\text{ and }x>1;\text{ find:}$
$\displaystyle \text{(i) }x-\frac{1}{x}\qquad \text{(ii) }x^3-\frac{1}{x^3}$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{x^2+1}{x}=3\frac{1}{3}=\frac{10}{3}$
$\displaystyle \therefore x+\frac{1}{x}=\frac{10}{3}$
$\displaystyle \text{(i) }\left(x-\frac{1}{x}\right)^2=\left(x+\frac{1}{x}\right)^2-4$
$\displaystyle =\left(\frac{10}{3}\right)^2-4$
$\displaystyle =\frac{100}{9}-\frac{36}{9}$
$\displaystyle =\frac{64}{9}$
$\displaystyle \text{Since }x>1,\ x-\frac{1}{x}>0$
$\displaystyle \therefore x-\frac{1}{x}=\frac{8}{3}$

$\displaystyle \text{(ii) }x^3-\frac{1}{x^3}=\left(x-\frac{1}{x}\right)^3+3\left(x-\frac{1}{x}\right)$
$\displaystyle =\left(\frac{8}{3}\right)^3+3\left(\frac{8}{3}\right)$
$\displaystyle =\frac{512}{27}+8$
$\displaystyle =\frac{512+216}{27}$
$\displaystyle =\frac{728}{27}$
$\displaystyle \therefore x^3-\frac{1}{x^3}=\frac{728}{27}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 15: }\text{The difference between two positive numbers is }4\text{ and the}$
$\displaystyle \text{difference between their cubes is }316.\text{ Find:}$
$\displaystyle \text{(i) their product}\qquad \text{(ii) the sum of their squares}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the two positive numbers be }a\text{ and }b,\text{ where }a>b.$
$\displaystyle a-b=4$
$\displaystyle a^3-b^3=316$
$\displaystyle a^3-b^3=(a-b)^3+3ab(a-b)$
$\displaystyle 316=4^3+3ab(4)$
$\displaystyle 316=64+12ab$
$\displaystyle 12ab=252$
$\displaystyle ab=21$
$\displaystyle \therefore \text{Their product is }21.$

$\displaystyle \text{(ii) }(a-b)^2=a^2+b^2-2ab$
$\displaystyle 4^2=a^2+b^2-2(21)$
$\displaystyle 16=a^2+b^2-42$
$\displaystyle a^2+b^2=58$
$\displaystyle \therefore \text{The sum of their squares is }58.$
$\displaystyle \\$

$\displaystyle \textbf{Exercise 4(E)}$

$\displaystyle \textbf{Question 1: }\text{Simplify:}$
$\displaystyle \text{(i) }(x+6)(x+4)(x-2)\qquad \text{(ii) }(x-6)(x-4)(x+2)$
$\displaystyle \text{(iii) }(x-6)(x-4)(x-2)\qquad \text{(iv) }(x+6)(x-4)(x-2)$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }(x+6)(x+4)(x-2)$
$\displaystyle =(x^2+10x+24)(x-2)$
$\displaystyle =x^3-2x^2+10x^2-20x+24x-48$
$\displaystyle =x^3+8x^2+4x-48$
$\displaystyle \therefore (x+6)(x+4)(x-2)=x^3+8x^2+4x-48.$

$\displaystyle \text{(ii) }(x-6)(x-4)(x+2)$
$\displaystyle =(x^2-10x+24)(x+2)$
$\displaystyle =x^3+2x^2-10x^2-20x+24x+48$
$\displaystyle =x^3-8x^2+4x+48$
$\displaystyle \therefore (x-6)(x-4)(x+2)=x^3-8x^2+4x+48.$

$\displaystyle \text{(iii) }(x-6)(x-4)(x-2)$
$\displaystyle =(x^2-10x+24)(x-2)$
$\displaystyle =x^3-2x^2-10x^2+20x+24x-48$
$\displaystyle =x^3-12x^2+44x-48$
$\displaystyle \therefore (x-6)(x-4)(x-2)=x^3-12x^2+44x-48.$

$\displaystyle \text{(iv) }(x+6)(x-4)(x-2)$
$\displaystyle =(x^2+2x-24)(x-2)$
$\displaystyle =x^3-2x^2+2x^2-4x-24x+48$
$\displaystyle =x^3-28x+48$
$\displaystyle \therefore (x+6)(x-4)(x-2)=x^3-28x+48.$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{Simplify using the following identity:}$
$\displaystyle (a\pm b)(a^2\mp ab+b^2)=a^3\pm b^3$
$\displaystyle \text{(i) }(2x+3y)(4x^2-6xy+9y^2)$
$\displaystyle \text{(ii) }\left(3x-\frac{5}{x}\right)\left(9x^2+15+\frac{25}{x^2}\right)$
$\displaystyle \text{(iii) }\left(\frac{a}{3}-3b\right)\left(\frac{a^2}{9}+ab+9b^2\right)$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }(2x+3y)(4x^2-6xy+9y^2)$
$\displaystyle =(2x+3y)\left[(2x)^2-(2x)(3y)+(3y)^2\right]$
$\displaystyle =(2x)^3+(3y)^3$
$\displaystyle =8x^3+27y^3$
$\displaystyle \therefore (2x+3y)(4x^2-6xy+9y^2)=8x^3+27y^3.$

$\displaystyle \text{(ii) }\left(3x-\frac{5}{x}\right)\left(9x^2+15+\frac{25}{x^2}\right)$
$\displaystyle =\left(3x-\frac{5}{x}\right)\left[(3x)^2+(3x)\left(\frac{5}{x}\right)+\left(\frac{5}{x}\right)^2\right]$
$\displaystyle =(3x)^3-\left(\frac{5}{x}\right)^3$
$\displaystyle =27x^3-\frac{125}{x^3}$
$\displaystyle \therefore \left(3x-\frac{5}{x}\right)\left(9x^2+15+\frac{25}{x^2}\right)=27x^3-\frac{125}{x^3}.$

$\displaystyle \text{(iii) }\left(\frac{a}{3}-3b\right)\left(\frac{a^2}{9}+ab+9b^2\right)$
$\displaystyle =\left(\frac{a}{3}-3b\right)\left[\left(\frac{a}{3}\right)^2+\left(\frac{a}{3}\right)(3b)+(3b)^2\right]$
$\displaystyle =\left(\frac{a}{3}\right)^3-(3b)^3$
$\displaystyle =\frac{a^3}{27}-27b^3$
$\displaystyle \therefore \left(\frac{a}{3}-3b\right)\left(\frac{a^2}{9}+ab+9b^2\right)=\frac{a^3}{27}-27b^3.$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{Using suitable identity, evaluate:}$
$\displaystyle \text{(i) }(104)^3\qquad \text{(ii) }(97)^3$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }(104)^3=(100+4)^3$
$\displaystyle \text{Using }(a+b)^3=a^3+3a^2b+3ab^2+b^3,$
$\displaystyle (104)^3=100^3+3(100)^2(4)+3(100)(4)^2+4^3$
$\displaystyle =10,00,000+1,20,000+4,800+64$
$\displaystyle =11,24,864$
$\displaystyle \therefore (104)^3=11,24,864.$

$\displaystyle \text{(ii) }(97)^3=(100-3)^3$
$\displaystyle \text{Using }(a-b)^3=a^3-3a^2b+3ab^2-b^3,$
$\displaystyle (97)^3=100^3-3(100)^2(3)+3(100)(3)^2-3^3$
$\displaystyle =10,00,000-90,000+2,700-27$
$\displaystyle =9,12,673$
$\displaystyle \therefore (97)^3=9,12,673.$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{Simplify:}$
$\displaystyle \frac{(x^2-y^2)^3+(y^2-z^2)^3+(z^2-x^2)^3}{(x-y)^3+(y-z)^3+(z-x)^3}$
$\displaystyle \text{Answer:}$
$\displaystyle (x^2-y^2)+(y^2-z^2)+(z^2-x^2)=0$
$\displaystyle \text{Using the identity, if }a+b+c=0,\text{ then }a^3+b^3+c^3=3abc,$
$\displaystyle (x^2-y^2)^3+(y^2-z^2)^3+(z^2-x^2)^3$
$\displaystyle =3(x^2-y^2)(y^2-z^2)(z^2-x^2)\qquad\ldots(1)$
$\displaystyle \text{Also, }(x-y)+(y-z)+(z-x)=0$
$\displaystyle \therefore (x-y)^3+(y-z)^3+(z-x)^3$
$\displaystyle =3(x-y)(y-z)(z-x)\qquad\ldots(2)$
$\displaystyle \therefore \frac{(x^2-y^2)^3+(y^2-z^2)^3+(z^2-x^2)^3}{(x-y)^3+(y-z)^3+(z-x)^3}$
$\displaystyle =\frac{3(x^2-y^2)(y^2-z^2)(z^2-x^2)}{3(x-y)(y-z)(z-x)}$
$\displaystyle =\frac{(x-y)(x+y)(y-z)(y+z)(z-x)(z+x)}{(x-y)(y-z)(z-x)}$
$\displaystyle =(x+y)(y+z)(z+x)$
$\displaystyle \therefore \text{The simplified expression is }(x+y)(y+z)(z+x).$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{Evaluate:}$
$\displaystyle \text{(i) }\frac{0.8\times0.8\times0.8+0.5\times0.5\times0.5}{0.8\times0.8-0.8\times0.5+0.5\times0.5}$
$\displaystyle \text{(ii) }\frac{1.2\times1.2\times1.2+0.3\times0.3\times0.3}{1.2\times1.2\times1.2-0.3\times0.3\times0.3}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) Let }a=0.8\text{ and }b=0.5.$
$\displaystyle \text{Using }a^3+b^3=(a+b)(a^2-ab+b^2),$
$\displaystyle \frac{0.8^3+0.5^3}{0.8^2-0.8\times0.5+0.5^2}$
$\displaystyle =\frac{(0.8+0.5)(0.8^2-0.8\times0.5+0.5^2)}{0.8^2-0.8\times0.5+0.5^2}$
$\displaystyle =0.8+0.5$
$\displaystyle =1.3$
$\displaystyle \therefore \text{The required value is }1.3.$

$\displaystyle \text{(ii) Let }a=1.2\text{ and }b=0.3.$
$\displaystyle \text{Using }a^3-b^3=(a-b)(a^2+ab+b^2),$
$\displaystyle \frac{1.2^3+0.3^3}{1.2^3-0.3^3}$
$\displaystyle =\frac{1.728+0.027}{1.728-0.027}$
$\displaystyle =\frac{1.755}{1.701}$
$\displaystyle =\frac{65}{63}$
$\displaystyle \therefore \text{The required value is }\frac{65}{63}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\text{If }a-2b+3c=0;\text{ state the value of }a^3-8b^3+27c^3.$
$\displaystyle \text{Answer:}$
$\displaystyle a-2b+3c=0$
$\displaystyle \text{Using the identity, if }x+y+z=0,\text{ then }x^3+y^3+z^3=3xyz.$
$\displaystyle \text{Taking }x=a,\ y=-2b\text{ and }z=3c,\text{ we get}$
$\displaystyle a^3+(-2b)^3+(3c)^3=3(a)(-2b)(3c)$
$\displaystyle a^3-8b^3+27c^3=-18abc$
$\displaystyle \therefore \text{The required value is }-18abc.$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: }\text{If }x+5y=10,\text{ find the value of}$
$\displaystyle x^3+125y^3+150xy-1000.$
$\displaystyle \text{Answer:}$
$\displaystyle x+5y=10$
$\displaystyle \text{Using }a^3+b^3=(a+b)^3-3ab(a+b),$
$\displaystyle x^3+125y^3=x^3+(5y)^3$
$\displaystyle =(x+5y)^3-3(x)(5y)(x+5y)$
$\displaystyle =(x+5y)^3-15xy(x+5y)$
$\displaystyle \therefore x^3+125y^3+150xy-1000$
$\displaystyle =(x+5y)^3-15xy(x+5y)+150xy-1000$
$\displaystyle =10^3-15xy(10)+150xy-1000$
$\displaystyle =1000-150xy+150xy-1000$
$\displaystyle =0$
$\displaystyle \therefore \text{The required value is }0.$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: }\text{If }x=3+2\sqrt2,\text{ find:}$
$\displaystyle \text{(i) }\frac{1}{x}\qquad \text{(ii) }x-\frac{1}{x}\qquad \text{(iii) }\left(x-\frac{1}{x}\right)^3\qquad \text{(iv) }x^3-\frac{1}{x^3}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }\frac{1}{x}=\frac{1}{3+2\sqrt2}$
$\displaystyle =\frac{1}{3+2\sqrt2}\times\frac{3-2\sqrt2}{3-2\sqrt2}$
$\displaystyle =\frac{3-2\sqrt2}{9-8}$
$\displaystyle =3-2\sqrt2$
$\displaystyle \therefore \frac{1}{x}=3-2\sqrt2.$

$\displaystyle \text{(ii) }x-\frac{1}{x}=(3+2\sqrt2)-(3-2\sqrt2)$
$\displaystyle =4\sqrt2$
$\displaystyle \therefore x-\frac{1}{x}=4\sqrt2.$

$\displaystyle \text{(iii) }\left(x-\frac{1}{x}\right)^3=(4\sqrt2)^3$
$\displaystyle =64\times2\sqrt2$
$\displaystyle =128\sqrt2$
$\displaystyle \therefore \left(x-\frac{1}{x}\right)^3=128\sqrt2.$

$\displaystyle \text{(iv) }x^3-\frac{1}{x^3}=\left(x-\frac{1}{x}\right)^3+3\left(x-\frac{1}{x}\right)$
$\displaystyle =128\sqrt2+3(4\sqrt2)$
$\displaystyle =128\sqrt2+12\sqrt2$
$\displaystyle =140\sqrt2$
$\displaystyle \therefore x^3-\frac{1}{x^3}=140\sqrt2.$
$\displaystyle \\$

$\displaystyle \textbf{Question 9: }\text{If }a+b=11\text{ and }a^2+b^2=65;\text{ find }a^3+b^3.$
$\displaystyle \text{Answer:}$
$\displaystyle a^2+b^2=(a+b)^2-2ab$
$\displaystyle 65=11^2-2ab$
$\displaystyle 65=121-2ab$
$\displaystyle 2ab=56$
$\displaystyle ab=28$
$\displaystyle a^3+b^3=(a+b)^3-3ab(a+b)$
$\displaystyle =11^3-3(28)(11)$
$\displaystyle =1331-924$
$\displaystyle =407$
$\displaystyle \therefore a^3+b^3=407.$
$\displaystyle \\$