If $a + b + c = 0$ , then prove that $a^{3} + b^{3} + c^{3} = 3 a b c$ .

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Solution

We know,
$a^{3} + b^{3} + c^{3} - 3 a b c = (a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - c a)$

Putting $a + b + c = 0$ on RHS, we get,

$a^{3} + b^{3} + c^{3} - 3 a b c = 0$

$⟹ a^{3} + b^{3} + c^{3} = 3 a b c$ , Hence proved.

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