Question

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

Answer 1

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

Applying componendo and dividendo:

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

Rearranging into algebraic cubic expansions:

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

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

Taking cube root on both sides:

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

Applying componendo and dividendo again:

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

$$\frac{2x}{4} = \frac{2y}{6}$$ 

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

Rearranging to find the ratio $$x : y$$:

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

$$x : y = 2 : 3$$