Class 8: Special Products and Expansions – Exercise 19D – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: Expand:

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

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

Answer:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\\$

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

Answer:

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

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

$\\$

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

Answer:

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

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

$\\$

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

Answer:

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

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

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

$\\$

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

Answer:

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

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

$\\$

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

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

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

Answer:

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

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

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

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

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

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

$\\$

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

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

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

Answer:

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

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

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

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

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

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

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

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

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

$\\$

Question 8: Evaluate the following:

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

Answer:

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

$= 1000000+27+92700$

$= 109272.7$

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

$= 1000000-8-58800$

$=941192$

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

$= 64964808$

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

$= 216000000-8-2152800$

$= 213847192$

$\\$

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

Answer:

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

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

$\\$

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

Answer:

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

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

$\\$

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

Answer:

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

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

$\\$

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

Answer:

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

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

$\\$

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

Answer:

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

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

$\\$

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

Answer:

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

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

$\\$

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

Answer:

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

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

$\\$

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

Answer:

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

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

ICSE / ISC / CBSE Mathematics Portal for K12 Students

Mathematics Made Easy for School Students

Class 8: Special Products and Expansions – Exercise 19D

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