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

Answer:

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}  

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 )

Answer:

\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)

Answer:

\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)

Answer:

\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)

Answer:

\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)  

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

Answer:

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)  

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

Answer:

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}  

\displaystyle \\

Question 8: Evaluate the following:

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

Answer:

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)

Answer:

\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 )

Answer:

\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 )

Answer:

\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)

Answer:

\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

Answer:

\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 )

Answer:

\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)

Answer:

\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)

Answer:

\displaystyle (4x-5z)^3=64x^3 -125x^3 -60xz (4x-5z)

\displaystyle \Rightarrow 64^3-125^3=23+60 \times 6 \times 2 = 728