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,

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

$\Rightarrow \frac{x^3+3{\cdot x}^2\cdot 2+3\cdot x\cdot 2^2+8}{x^3-3\cdot x^2\cdot 2+3\cdot x\cdot 2^2-8}=\frac{y^3+3\cdot y^2\cdot 3+3\cdot y\cdot 3^2+27}{y^3-3\cdot y^2\cdot 3+3\cdot y\cdot 3^2-27}$

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

Taking cube roots both sides of equation,

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

Applying componendo and dividendo again,

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

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

$\Rightarrow \frac{x}{y}=\frac{2}{3}$ i.e. $x:y=2:3$