If $16\Big(\dfrac{a - x}{a + x}\Big)^{3} = \Big(\dfrac{a + x}{a - x}\Big)$, prove that x = $\dfrac{a}{3}$.

Given,

$\Rightarrow 16\Big(\dfrac{a - x}{a + x}\Big)^3 = \Big(\dfrac{a + x}{a - x}\Big)$

Let, $r = \dfrac{a + x}{a - x} , \dfrac{1}{r} = \dfrac{a - x}{a + x}.$

Substituting value of r and $\dfrac{1}{r}$ in $16\Big(\dfrac{a - x}{a + x}\Big)^3 = \Big(\dfrac{a + x}{a - x}\Big)$, we get :

$\Rightarrow 16 \times \Big(\dfrac{1}{r}\Big)^3 = r \\[1em] \Rightarrow 16 = r^4 \\[1em] \Rightarrow r^4 = 2^4 \\[1em] \Rightarrow r = 2. \\[1em] \Rightarrow \dfrac{a + x}{a - x} = 2 \\[1em] \Rightarrow a + x = 2a - 2x \\[1em] \Rightarrow 3x = a \\[1em] \Rightarrow x = \dfrac{a}{3}.$

Hence, proved that x = $\dfrac{a}{3}$.