Factorise:
$a^{3} + 3 a^{2} b + 3 a b^{2} + 2 b^{3}$

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Solution

We have,
$a^{3} + {3 a}^{2} b + 3 {a b}^{2} + 2 b^{3}$
$= a^{3} + b^{3} + 3 a b (a + b) + b^{3}$
$= {(a + b)}^{3} + b^{3}$
$= (a + b + b) ({(a + b)}^{2} + b^{2} - b (a + b))$ [Using $x^{3} + y^{3} = x + y (x^{2} + x y + y^{2})$ ]
$= (a + 2 b) (a^{2} + b^{2} + 2 a b + b^{2} - a b - b^{2})$
$= (a + 2 b) (a^{2} + a b + b^{2})$

Hence, this is the answer.

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