Question
Factorize the following expressions: $(a + b)^3 - 8(a - b)^3$
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Answer
$=(a+b)^3-[2(a-b)]^3=(a+b)^3-[2 a-2 b]^3$
$=(a+b-(2 a-2 b))\left((a+b)^2+(a+b)(2 a-2 b)+(2 a-2 b)^2\right)$
$\therefore\left[a^3-b^3=(a-b)\left(a^2+a b+b^2\right)\right]$
$=(a+b-2 a+2 b)\left(a^2+b^2+2 a b+(a+b)(2 a-2 b)+(2 a-2 b)^2\right)$
$=(a+b-2 a+2 b)\left(a^2+b^2+2 a b+2 a^2-2 a b+2 a b-2 b^2+(2 a-2 b)^2\right)$
$=(3 b-a)\left(3 a^2+2 a b-b^2+(2 a-2 b)^2\right)$
$=(3 b-a)\left(3 a^2+2 a b-b^2+4 a^2+4 b^2-8 a b)=(3 b-a)(3 a^2+4 a^2-b^2+4 b^2-8 a b+2 a b\right)$
$=(3 b-a)\left(7 a^2+3 b^2-6 a b\right)$
$\therefore(a+b)^3-8(a-b)^3=(3 b-a)\left(7 a^2+3 b^2-6 a b\right)$
$=(a+b-(2 a-2 b))\left((a+b)^2+(a+b)(2 a-2 b)+(2 a-2 b)^2\right)$
$\therefore\left[a^3-b^3=(a-b)\left(a^2+a b+b^2\right)\right]$
$=(a+b-2 a+2 b)\left(a^2+b^2+2 a b+(a+b)(2 a-2 b)+(2 a-2 b)^2\right)$
$=(a+b-2 a+2 b)\left(a^2+b^2+2 a b+2 a^2-2 a b+2 a b-2 b^2+(2 a-2 b)^2\right)$
$=(3 b-a)\left(3 a^2+2 a b-b^2+(2 a-2 b)^2\right)$
$=(3 b-a)\left(3 a^2+2 a b-b^2+4 a^2+4 b^2-8 a b)=(3 b-a)(3 a^2+4 a^2-b^2+4 b^2-8 a b+2 a b\right)$
$=(3 b-a)\left(7 a^2+3 b^2-6 a b\right)$
$\therefore(a+b)^3-8(a-b)^3=(3 b-a)\left(7 a^2+3 b^2-6 a b\right)$
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