Question
If $a + 2b + c = 0$; then show that $a^3 + 8b^3 + c^3 = 6abc$
✓
Answer
$a + 2b + c = 0 \dots...(i)$
$\Rightarrow (a + 2b) + c = 0$
$\Rightarrow (a + 2b)^3 + c^3+ 3(a + 2b) c(a + 2b + c) = 0$
$\Rightarrow a^3+ 8b^2+ 6ab (a + 2b) + c^3+ 0 = 0$
$\Rightarrow a^3 + 8b^3 + c^3+ 6ab (a + 2b) = 0 \dots....(2)$
Using $(1),$ we get $a + 2b = -c$
From $(2),$
$a^3 + 8b^3+ 6ab (-c) = 0$
$\Rightarrow a^3 + 8b^3+ c^3 = 6abc.$
$\Rightarrow (a + 2b) + c = 0$
$\Rightarrow (a + 2b)^3 + c^3+ 3(a + 2b) c(a + 2b + c) = 0$
$\Rightarrow a^3+ 8b^2+ 6ab (a + 2b) + c^3+ 0 = 0$
$\Rightarrow a^3 + 8b^3 + c^3+ 6ab (a + 2b) = 0 \dots....(2)$
Using $(1),$ we get $a + 2b = -c$
From $(2),$
$a^3 + 8b^3+ 6ab (-c) = 0$
$\Rightarrow a^3 + 8b^3+ c^3 = 6abc.$
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