If, $\dfrac{x^{3} + 3x}{3x^{2} + 1} = \dfrac{341}{91}$, prove that x = 11.

Given,

$\dfrac{x^{3} + 3x}{3x^{2} + 1} = \dfrac{341}{91}$

Solving L.H.S:

Applying Componendo and Dividendo, we get :

$\Rightarrow \dfrac{(x^{3} + 3x) + (3x^{2} + 1)}{(x^{3} + 3x) - (3x^{2} + 1)} \\[1em] \Rightarrow \dfrac{(x^{3} + 3x + 3x^{2} + 1)}{(x^{3} + 3x - 3x^{2} - 1)} \\[1em] \Rightarrow \dfrac{(x + 1)^3}{(x - 1)^3} \\[1em] \Rightarrow \Big(\dfrac{x + 1}{x - 1}\Big)^3.$

Solving R.H.S:

Apply Componendo and Dividendo:

$\Rightarrow \dfrac{341 + 91}{341 - 91} \\[1em] \Rightarrow \dfrac{432}{250}.$

Equating L.H.S. and R.H.S.,

$\Rightarrow \Big(\dfrac{x + 1}{x - 1}\Big)^3 = \dfrac{432}{250} \\[1em] \Rightarrow \Big(\dfrac{x + 1}{x - 1}\Big)^3 = \dfrac{216}{125} \\[1em] \Rightarrow \Big(\dfrac{x + 1}{x - 1}\Big)^3 = \Big(\dfrac{6}{5}\Big)^3 \\[1em] \Rightarrow \Big(\dfrac{x + 1}{x - 1}\Big) = \Big(\dfrac{6}{5}\Big) \\[1em] \Rightarrow 5(x + 1) = 6(x - 1) \\[1em] \Rightarrow 5x + 5 = 6x - 6 \\[1em] \Rightarrow 6x - 5x = 6 + 5 \\[1em] \Rightarrow x = 11.$

Hence, proved that x = 11.