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

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

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`(x^3+12x)/(6x^2+8) = (y^3+27y)/(9y^2+27)`
Using componendo and dividendo rule both sides,
`=>(x^3+12x+6x^2+8)/(x^3+12x-6x^2-8)= (y^3+27y+9y^2+27)/(y^3+27y-9y^2-27)`
`=>(x^3+2^3+3(x)(2)(x+2))/(x^3-2^3-3(x)(2)(x-2))= (y^3+3^3+3(y)(3)(y+3))/(y^3-3^3-3(y)(3)(x-3))`
`=>((x+2)^3)/((x-2)^3) = ((y+3)^3)/((y-3)^3) `
`=>((x+2)/(x-2))^3 = ((y+3)/(y-3))^3`
Applying cube roots on both sides,
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