Question
If $a + b + c = 0$; then show that $a^3 + b^3 + c^3 = 3abc$.
✓
Answer
$a + b + c = 0 \dots...(i)$
$\Rightarrow (a + b) + c = 0$
Cubing both sides
$\Rightarrow (a + b)^3 + c^3 + 3(a + b) (c) (a+ b + c) = 0$
$\Rightarrow a^3 + b^3+ 3ab (a + b) + c^3 + 0 = 0$
$\Rightarrow a^3 + b^3+ c^3 + 3ab (a + b) = 0 \dots...(2)$
Using $(i),$ we get,
$a + b = -c$ From $(2),$
$a^3 + b^3 + c^3 + 3ab (-c) = 0$
$\Rightarrow a^3 + b^3 + c^3 = 3abc$.
$\Rightarrow (a + b) + c = 0$
Cubing both sides
$\Rightarrow (a + b)^3 + c^3 + 3(a + b) (c) (a+ b + c) = 0$
$\Rightarrow a^3 + b^3+ 3ab (a + b) + c^3 + 0 = 0$
$\Rightarrow a^3 + b^3+ c^3 + 3ab (a + b) = 0 \dots...(2)$
Using $(i),$ we get,
$a + b = -c$ From $(2),$
$a^3 + b^3 + c^3 + 3ab (-c) = 0$
$\Rightarrow a^3 + b^3 + c^3 = 3abc$.
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