Factorise: $a^{3} + b^{3} + c^{3} - 3 a b c$

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Solution

To factorise: $a^{3} + b^{3} + c^{3} - 3 a b c$

From the standard identity,

$a^{3} + b^{3} + c^{3} = (a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - c a) + 3 a b c$
$\Rightarrow a^{3} + b^{3} + c^{3} - 3 a b c = (a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - c a)$

Hence, the factorised form is,
$(a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - c a)$

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