If $x = -2$ is the common solution of quadratic equations $ax^2 + x - 3a = 0$ and $x^2 + bx + b = 0$, then find the value of $a^2b$.

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Step 1: $x = -2$ is the common solution of $ax^2 + x - 3a = 0$ and $x^2 + bx + b = 0$.
$	herefore a(-2)^2 + (-2) - 3a = 0 \Rightarrow 4a - 2 - 3a = 0$
$a = 2$
Step 2: And $(-2)^2 + b(-2) + b = 0 \Rightarrow 4 - 2b + b = 0 \Rightarrow b = 4$
Step 3: $a^2b = 4 	imes 4 = 16$

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