Factorize $a^{3} - 3 a^{2} b + 3 a b^{2} - b^{3} + 8$

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Solution

We have, $a^{3} - 3 a^{2} b + 3 a b^{2} - b^{3} + 8$
$= (a^{3} - 3 a^{2} b + 3 a b^{2} - b^{3}) + 8$ $[∵ (a - b)^{3} = a^{3} - 3 a^{2} b + 3 a b^{2} - b^{3}]$
$= (a - b)^{3} + (2)^{3}$
$= [(a - b) + 2] [(a - b)^{2} - (a - b) \times 2 + (2)^{2}]$ $[∵ a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})]$
$= (a - b + 2) (a^{2} + b^{2} - 2 a b - 2 a + 2 b + 4)$ $[∵ (a - b)^{2} = a^{2} + b^{2} - 2 a b]$

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Similar questions

a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3} - 8

(a + b)^{3} = a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}

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

Which of the following is an expansion of the identity ${(a - b)}^{3}$ ?

(a + b)^{3} = a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}

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