Question

Given `(x^3 + 12x)/(6x^2 + 8) = (y^3 + 27y)/(9y^2 + 27)`  using componendo and divendo find x : y

Answer 1

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

Applying componenedo and divendendo we get

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

`=> (x^3 + 3(1)(4)x + 3(1)(2)x^2 + 2^3)/(x^3 + 3(1)(4)x - 3(1)(2)x^2 - 2^3) = (y^3 + 3(1)(9)y + 3(1)(3)y^2 + 3^3)/(y^3 + 3(1)(9)y - 3(1)(3)y^2 - 3^3)`

`=> (x^3 + 3(1)(4)x + 3(1)(2)x^2 + 2^3)/(x^3 - 3(1)(2)x^2 + 3(1)(4)x - 2^3) =(y^3 + 3(1)(9)y + 3(1)(3)y^2 + 3^3)/(y^3 - 3(1)(3)y^2 + 3(1)(9)y - 3^3)`

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

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

Again applying componendo and dividendo we get

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

`=> (2x)/4 = (2y)/6`

`=> x/2 = y/3`

Applying alternendo we get

`x/y = 2/3`

=> x : y = 2  : 3