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