Given $\dfrac{x^3 + 12x}{6x^2 + 8} = \dfrac{y^3 + 27y}{9y^2 + 27}$. Using componendo and dividendo find x : y.

Given,

$\dfrac{x^3 + 12x}{6x^2 + 8} = \dfrac{y^3 + 27y}{9y^2 + 27}$

Applying componendo and dividendo we get,

$\Rightarrow \dfrac{x^3 + 12x + 6x^2 + 8}{x^3 + 12x - 6x^2 - 8} = \dfrac{y^3 + 27y + 9y^2 + 27}{y^3 + 27y - 9y^2 - 27} \\[1em] \Rightarrow \dfrac{(x + 2)^3}{(x - 2)^3} = \dfrac{(y + 3)^3}{(y - 3)^3} \\[1em] \Rightarrow \dfrac{x + 2}{x - 2} = \dfrac{y + 3}{y - 3}$

Applying componendo and dividendo again we get,

$\Rightarrow \dfrac{x + 2 + x - 2}{x + 2 - (x - 2)} = \dfrac{y + 3 + y - 3}{y + 3 - (y - 3)} \\[1em] \Rightarrow \dfrac{2x}{4} = \dfrac{2y}{6} \\[1em] \Rightarrow \dfrac{x}{2} = \dfrac{y}{3} \\[1em] \Rightarrow \dfrac{x}{y} = \dfrac{2}{3} \\[1em] \Rightarrow x : y = 2 : 3.$

Hence, x : y = 2 : 3.