Question 1: Expand:

i) $\displaystyle (2a+3b+4c)^2$ ii) $\displaystyle (a+2b-5c)^2$ iii) $\displaystyle (3x +2y-z)^2$

iv) $\displaystyle (3x-2y-1)^2$ v) $\displaystyle \Big( \frac{a}{2} - \frac{b}{3} + \frac{c}{4} \Big)^2$ vi) $\displaystyle \Big(x- \frac{2}{x} +3 \Big)^2$

i) $\displaystyle (2a+3b+4c)^2$

$\displaystyle = (4a^2 +9b^(2 )+16c^2) +2 (6ab+12bc+8ac)$

$\displaystyle = 4a^(2 )+9b^2+16c^2+12ab+24bc+16ac$

ii) $\displaystyle (a+2b-5c)^2$

$\displaystyle = (a^2+4b^2+25c^2) +2 (2ab -10bc-5ac)$

$\displaystyle = a^2+4b^2+25c^2+4ab-20bc-10ac$

iii) $\displaystyle (3x +2y-z)^2$

$\displaystyle = (9x^2+4y^2+z^2 ) +2 (6xy-2yz-6xz)$

$\displaystyle = 9x^2+4y^2+z^2+12xy-4yz-12xz$

iv) $\displaystyle (3x-2y-1)^2$

$\displaystyle = (9x^2+4y^2+1)+ 2 (-6xy+2y-3x)$

$\displaystyle = 9x^2+4y^2+1-12xy+4y-6x$

v) $\displaystyle \Big( \frac{a}{2} - \frac{b}{3} + \frac{c}{4} \Big)^2$

$\displaystyle = \Big( \frac{a^2}{4} + \frac{b^2}{9} + \frac{c^2}{16} \Big)+2 \Big( \frac{-ab}{6} - \frac{bc}{12} + \frac{ac}{8} \Big)$

$\displaystyle = \Big( \frac{a^2}{4} + \frac{b^2}{9} + \frac{c^2}{16} \Big)- \frac{ab}{3} - \frac{bc}{6} + \frac{ac}{4}$Â

vi) $\displaystyle \Big (x- \frac{2}{x} +3 \Big)^2$

$\displaystyle = x^2 + \frac{4}{x^2} +9-2\times \Big(-2- \frac{6}{x} +3x \Big)$

$\displaystyle = x^2+ \frac{4}{x^2} +9+4+ \frac{12}{x} -6x$

$\displaystyle = x^2+ \frac{4}{x^2} + \frac{12}{x} -6x+13$

$\displaystyle \\$

Question 2: If $\displaystyle (a+b+c)=10$ and $\displaystyle ab+bc+ac = 31$ Find $\displaystyle (a^2+b^2+c^2 )$

$\displaystyle (a+b+c)^2=(a^2+b^2+c^2 )+2(ab+bc+ca)$

$\displaystyle \Rightarrow (a^2+b^2+c^2 )=10^2-2\times 3=100-62=38$

$\displaystyle \\$

Question 3: If $\displaystyle ((a+b+c)=9$ and $\displaystyle (a^2+b^2+c^2 )=29$ find $\displaystyle (ab+bc+ac)$

$\displaystyle (a+b+c)^2=(a^2+b^2+c^2 )+2(ab+bc+ac)$

$\displaystyle \Rightarrow (ab+bc+ac)= \frac{1}{2} (9^2-29)= \frac{(81-29)}{2} = 26$

$\displaystyle \\$

Question 4: If $\displaystyle (a^2+b^2+c^2 )=45$ and $\displaystyle (ab+bc+ac)=38$ find$\displaystyle (a+b+c)$

$\displaystyle (a+b+c)^2=(a^2+b^2+c^2 )+2(ab+bc+ac)$

$\displaystyle \Rightarrow (a+b+c)^2=45+2\times 38=121$

$\displaystyle \Rightarrow (a+b+c)= \pm 11$

$\displaystyle \\$

Question 5: If $\displaystyle (a+b-c)= 11$ and $\displaystyle (a^2+b^2+c^2 )=89$ find $\displaystyle (ab-bc-ca)$

$\displaystyle (a+b-c)^2=(a^2+b^2+c^2 )+2(ab-bc-ac)$

$\displaystyle \Rightarrow (ab-bc-ca)= \frac{1}{2} (81-11^2 )=-20$

$\displaystyle \\$

Question 6: Expanding using Formula $\displaystyle (a+b)^3=a^3+b^3+3ab(a+b)$

i) $\displaystyle (x+3)^3= x^3+27+9x(x+3)$ ii) $\displaystyle (4+a)^3=64+a^3+3\times4a (4+a)$Â

iii) $\displaystyle (6a+5b)^3= (6a)^3+(5b)^3+3(6a)(5b)(6a+5b)$

i) $\displaystyle (x+3)^3= x^3+27+9x(x+3)$

$\displaystyle = x^3+9x^2+27x+27$

ii) $\displaystyle (4+a)^3=64+a^3+3\times 4a (4+a)$

$\displaystyle = a^3+12a^2+12a+64$

iii) $\displaystyle (6a+5b)^3= (6a)^3+(5b)^3+3(6a)(5b)(6a+5b)$

$\displaystyle =216a^3+125b^3+540a^2 b+450ab^2$

$\displaystyle \\$

Question 7: Expand Unity Formula $\displaystyle (a-b)^3=a^3-b^3-3ab(a-b)$

i) $\displaystyle (x-4)^3=x^3-64-12x(x-4)$ ii) $\displaystyle (2-y)^3=8-y^3-4y(2-y)$Â

iii) $\displaystyle (4x-5y)^3=64 x^3-125y^3-60xy(4x-5y)$ iv) $\displaystyle \Big( x - \frac{1}{x} \Big)^3$

i) $\displaystyle (x-4)^3=x^3-64-12x(x-4)$

$\displaystyle = x^3-12x^2+48x-64$

ii) $\displaystyle (2-y)^3=8-y^3-4y(2-y)$

$\displaystyle = -y^3+4y-8y+8$

iii) $\displaystyle (4x-5y)^3=64 x^3-125y^3-60xy(4x-5y)$

$\displaystyle =64x^3-125y^3-240x^2 y+300 xy^2$

iv) $\displaystyle \Big( x - \frac{1}{x} \Big)^3$

$\displaystyle = x^3 - \frac{1}{x^3} - 3 \Big( x - \frac{1}{x} \Big)$

$\displaystyle = x^3 - \frac{1}{x^3} - 3x + \frac{3}{x}$Â

$\displaystyle \\$

Question 8: Evaluate the following:

i) $\displaystyle (103)^3$ ii) $\displaystyle (98)^3$ iii) $\displaystyle (402)^3$ iv) $\displaystyle (598)^3$

i) $\displaystyle (103)^3=(100+3)^3=1000000+27+900(100+3)$

$\displaystyle = 1000000+27+92700$

$\displaystyle = 109272.7$

ii) $\displaystyle (98)^3=(100-2)^3=1000000-8-600(100-2)$

$\displaystyle = 1000000-8-58800$

$\displaystyle =941192$

iii) $\displaystyle (402)^3=(400+2)^3=64000000+8+2400(400+2)$

$\displaystyle = 64964808$

iv) $\displaystyle (598)^3=(600-2)^(3 )=216000000-8-3600(600-2)$

$\displaystyle = 216000000-8-2152800$

$\displaystyle = 213847192$

$\displaystyle \\$

Question 9: If $\displaystyle \Big(a+ \frac{1}{a} \Big)=6,$ find the value of $\displaystyle \Big(a^3+ \frac{1}{a^3} \Big)$

$\displaystyle (a+\frac{1}{a})^3= a^3+ \frac{1}{a^3} +3 (a+ \frac{1}{a} )$

$\displaystyle \Rightarrow a^3+ \frac{1}{a^3} =6^3-36=198$

$\displaystyle \\$

Question 10: If $\displaystyle \Big(2a + \frac{1}{2a} \Big)=4$ Find the value of $\displaystyle \Big( 8a^3+ \frac{1}{8a^3} \Big )$

$\displaystyle (2a + \frac{1}{2a} )^3=8a^3+ \frac{1}{8a^3} +3(2a + \frac{1}{2a} )$

$\displaystyle \Rightarrow 8a^3+ \frac{1}{8a^3} =4^3-3\times 4=64-12=52$

$\displaystyle \\$

Question 11: If $\displaystyle \Big(x- \frac{1}{x} \Big)=7$ Find the value of $\displaystyle \Big(x^3- \frac{1}{x^3} \Big )$

$\displaystyle (x- \frac{1}{x} )^3=x^3- \frac{1}{x^3} -3(x- \frac{1}{x} )$

$\displaystyle \Rightarrow x^3- \frac{1}{x^3} =7^3+3\times 7=343 +21=364$

$\displaystyle \\$

Question 12: If $\displaystyle \Big(3p- \frac{1}{3p} \Big)=5,$ Find the value of $\displaystyle \Big(27p^3- \frac{1}{27 p^3} \Big)$

$\displaystyle (3p- \frac{1}{3p} )^3= (27 p^3- \frac{1}{27 p^3} )-3 (3p- \frac{1}{3p} )$

$\displaystyle \Rightarrow (27p^3- \frac{1}{27 p^3} )=5^3+3\times 5=125 + 15 =140$

$\displaystyle \\$

Question 13: If $\displaystyle a + b = 9$ and $\displaystyle ab=20$ find the value of $\displaystyle a^3+b^3$

$\displaystyle (a+b)^3 = a^3+b^3+3ab(a+b)$

$\displaystyle \Rightarrow a^3+ b^3= 9^3-3\times 20\times 9=189$

$\displaystyle \\$

Question 14: If $\displaystyle (2x + 3y) = 7$ and $\displaystyle xy=2$ find the value of $\displaystyle (8x^3+27y^3 )$

$\displaystyle (2x+3y)^3= 8x^3+27y^3+18 xy (2x+3y)$

$\displaystyle \Rightarrow 8x^3+27y^3=7^3-18\times 2\times 7=91$

$\displaystyle \\$

Question 15: If $\displaystyle (a-b)=5$ and $\displaystyle ab=14$ Find the value of $\displaystyle (a^3-b^3)$

$\displaystyle (a-b)^3 = (a^3-b^3 )-3ab(a-b)$

$\displaystyle \Rightarrow a^3-b^3=5^3-3\times 5\times 14=-85$

$\displaystyle \\$

Question 16: If $\displaystyle (4x-5z)=2$ and $\displaystyle xz=6$ find the value of $\displaystyle (64x^3 -125z^3)$

$\displaystyle (4x-5z)^3=64x^3 -125x^3 -60xz (4x-5z)$
$\displaystyle \Rightarrow 64^3-125^3=23+60 \times 6 \times 2 = 728$