Question

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

Answer 1

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

Applying componendo and dividendo

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

$\frac{{(x+2)}^3}{{(x-2)}^3}=\frac{{(y+3)}^3}{{(y-3)}^3}$

$\frac{x+2}{x-2}=\frac{y+3}{y-3}$

Again, applying componendo and dividendo

$\frac{x+2+x-2}{x+2-x+2}=\frac{y+3+y-3}{y+3-y+3}$

$\frac{2x}{4}=\frac{2y}{6}$

$\frac{x}{y}=\frac{4}{6}=\frac{2}{3}$

$x:y=2:3$.