Using properties of proportion solve the following: $\;$ $\dfrac{x^3 + 3x}{3x^2 + 1} = \dfrac{341}{91}$
Given: $\;$ $\dfrac{x^3 + 3x}{3x^2 + 1} = \dfrac{341}{91}$
i.e. $\;$ $\dfrac{x^3 + 3x + 3x^2 + 1}{x^3 + 3x - 3x^2 - 1} = \dfrac{341 + 91}{341 - 91}$ $\;\;\;$ [By componendo-dividendo]
i.e. $\;$ $\dfrac{\left(x + 1\right)^3}{\left(x - 1\right)^3} = \dfrac{432}{250}$
i.e. $\;$ $\dfrac{\left(x + 1\right)^3}{\left(x - 1\right)^3} = \dfrac{216}{125}$
i.e. $\;$ $\dfrac{\left(x + 1\right)^3}{\left(x - 1\right)^3} = \dfrac{6^3}{5^3}$
i.e. $\;$ $\dfrac{x + 1}{x - 1} = \dfrac{6}{5}$
i.e. $\;$ $\dfrac{x + 1 + x - 1}{x + 1 - x + 1} = \dfrac{6 + 5}{6 - 5}$ $\;\;\;$ [By componendo-dividendo]
i.e. $\;$ $\dfrac{2x}{2} = \dfrac{11}{1}$
$\implies$ $x = 11$