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If $a x^{2} + 2 a^{2} x + b^{3}$ is divisible by $x + a$ , then

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Both A, B

$a^{2} + a b + b^{2} = 0$

$a = b$

answer is C.

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Detailed Solution

$p (x) = a x^{2} + 2 a^{2} x + b^{3}$

$p (x)$ is divisible by $(x + a)$

$\begin{array}{l} p (- a) = 0 \\ a {(- a)}^{2} + 2 a^{2} (- a) + b^{3} = 0 \\ a^{3} - 2 a^{3} + b^{3} = 0 \\ b^{3} - a^{3} = 0 \\ (b - a) (b^{2} + a^{2} + a b) = 0 \\ \Rightarrow b = a or a^{2} + a b + b^{2} = 0 \end{array}$

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