Question

Given `(x^3 + 12x)/(6x^2 + 8) = (y^3 + 27y)/(9y^2 + 27)`. Using componendo and dividendo, find x : y.

Answer 1

`(x^3 + 12x)/(6x^2 + 8) = (y^3 + 27y)/(9y^2 + 27)`

Applying componendo and dividendo, we get 

`((x^3 + 12x) + (6x^2 + 8))/((x^3 + 12x) - (6x^2 + 8)) = ((y^3 + 27y) + (9y^2 + 27))/((y^3 + 27y) - (9y^2 + 27))`

`\implies ((x^3 + 12x) + (6x^2 + 8))/((x^3 + 12x) - (6x^2 - 8)) = ((y^3 + 27y) + (9y^2 + 27))/((y^3 + 27y) - (9y^2 - 27))`

`\implies (x^3 + 6x(2 + x) + 2^3)/(x^3 - 6x (x - 2) - 2^3) = ((y^3 + 9y)(y + 3) + 3^3)/((y^3 - 9y) (y - 3) - 3^3)`

`\implies (x + 2)^3/(x - 2)^3 = (y + 3)^3/(y - 3)^3`

Again applying componendo and dividendo, we get

`(x + 2 + x - 2)/(x + 2 - x + 2) = (y + 3 + y - 3)/(y + 3 - y + 3)`

`\implies (2x)/4 = (2y)/6`

`\implies x/2 = y/3`

Applying alternendo, we get

`\implies x/y = 2/3`

∴ x : y = 2 : 3