Question
Factorize:
$a^3-3 a^2 b+3 a b^2-b^3+8$
$a^3-3 a^2 b+3 a b^2-b^3+8$
✓
Answer
$a^3-3 a^2 b+3 a b^2-b^3+8$
$\left.=(a-b)^3+8\right)\left[\because a^3-b^3+3 a^2 b+3 a b^2=(a-b)^3\right]$
$=(a-b)^3+2^3$
$=(a-b+2)\left((a-b)^2-(a-b)^2+2^2\right)\left[\because a^3+b^3=(a+b)\left(a^2-a b+b^2\right)\right]$
$=(a-b+2)\left(a^2+b^2-2 a b-2 a+2 b+4\right)$
$\therefore a^3-3 a^2 b+3 a b^2-b^3+8$
$=(a-b+2)\left(a^2+b^2-2 a b-2 a+2 b+4\right)$
$\left.=(a-b)^3+8\right)\left[\because a^3-b^3+3 a^2 b+3 a b^2=(a-b)^3\right]$
$=(a-b)^3+2^3$
$=(a-b+2)\left((a-b)^2-(a-b)^2+2^2\right)\left[\because a^3+b^3=(a+b)\left(a^2-a b+b^2\right)\right]$
$=(a-b+2)\left(a^2+b^2-2 a b-2 a+2 b+4\right)$
$\therefore a^3-3 a^2 b+3 a b^2-b^3+8$
$=(a-b+2)\left(a^2+b^2-2 a b-2 a+2 b+4\right)$
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