If $2 a + 3 b + c = 0$ then show that : $8 a^{3} + 27 b^{3} + c^{3} = 18 a b c$

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Solution

Given: $2 a + 3 b + c = 0$
$\Rightarrow 2 a + 3 b = - c$
Cubing both sides, we get
${(2 a + 3 b)}^{3} = {(- c)}^{3}$
$\Rightarrow 8 a^{3} + 27 b^{3} + 18 a b (2 a + 3 b) = - c^{3}$
$\Rightarrow 8 a^{3} + 27 b^{3} + 18 a b (- c) = - c^{3}$ since $2 a + 3 b = - c$
$\Rightarrow 8 a^{3} + 27 b^{3} - 18 a b c = - c^{3}$
$\Rightarrow 8 a^{3} + 27 b^{3} + c^{3} = 18 a b c$
Hence proved.

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