Find the value

Question:

Find the value

$(a+b)^{3}-8(a-b)^{3}$

Solution:

$=(a+b)^{3}-[2(a-b)]^{3}$

$=(a+b)^{3}-[2 a-2 b]^{3}$

$=(a+b-(2 a-2 b))\left((a+b)^{2}+(a+b)(2 a-2 b)+(2 a-2 b)^{2}\right)$

$\therefore\left[a^{3}-b^{3}=(a-b)\left(a^{2}+a b+b^{2}\right)\right]$

$=(a+b-2 a+2 b)\left(a^{2}+b^{2}+2 a b+(a+b)(2 a-2 b)+(2 a-2 b)^{2}\right)$

$=(a+b-2 a+2 b)\left(a^{2}+b^{2}+2 a b+2 a^{2}-2 a b+2 a b-2 b^{2}+(2 a-2 b)^{2}\right)$

$=(3 b-a)\left(3 a^{2}+2 a b-b^{2}+(2 a-2 b)^{2}\right)$

$=(3 b-a)\left(3 a^{2}+2 a b-b^{2}+4 a^{2}+4 b^{2}-8 a b\right)$

$=(3 b-a)\left(3 a^{2}+4 a^{2}-b^{2}+4 b^{2}-8 a b+2 a b\right)$

$=(3 b-a)\left(7 a^{2}+3 b^{2}-6 a b\right)$

$\therefore(a+b)^{3}-8(a-b)^{3}=(3 b-a)\left(7 a^{2}+3 b^{2}-6 a b\right)$

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