Given, $\frac{a + 2b + x}{a - 2b + x} = \frac{(a + b)^2}{(a - b)^2}, a \neq 0$
Using properties of proportion, find the value of $x$.
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Given,
$\frac{a + 2b + x}{a - 2b + x} = \frac{(a + b)^2}{(a - b)^2}$
Applying componendo and dividendo, we get :
$$\begin{aligned}\Rightarrow \frac{(a + 2b + x) + (a - 2b + x)}{(a + 2b + x) - (a - 2b + x)} = \frac{(a + b)^2 + (a - b)^2}{(a + b)^2 - (a - b)^2} \\[1em] \Rightarrow \frac{2a + 2x}{4b} = \frac{2(a^2 + b^2)}{4ab} \\[1em] \Rightarrow \frac{2(a + x)}{4b} = \frac{2(a^2 + b^2)}{4ab} \\[1em] \Rightarrow a + x = \frac{a^2 + b^2}{a} \\[1em] \Rightarrow x = \frac{a^2 + b^2}{a} - a \\[1em] \Rightarrow x = \frac{a^2 + b^2 - a^2}{a} \\[1em] \Rightarrow x = \frac{b^2}{a}.\end{aligned}$$
Hence, $\bm{x} = \mathbf{\frac{b^2}{a}}$.