If (x+y) `prop` (x-y) , then show that `(x^(3) +y^(3)) prop (x^(3)-y^(3))`.

Explanation

(x+y) `prop` (x-y) `rArr (x+y)` =k (x-y) (where k `ne` 0= variation constant )
`rArr (x+y)/(x-y)=k `
`rArr (x+y+x-y)/(x+y-x+y)=(k+1)/(k-1)` [by componendo and dividendo]
`rArr (2x)/(2y)=(k+1)/(k-1)rArr(x)/(y)=(k+1)/(k-1)=m("let") [when (k+1)/(k-1)=m]`
`rArr (x/y)^(3) =m^(3) ` (cubing both the sides )
`rArr x^(3)/y^(3)=m^(3)rArr (x^(3)+y^(3))/(x^(3)-y^(3))=(m^(3)+1)/(m^(3)-1)` [by componendo and dividendo]
`(x^(3)+y^(3))/(x^(3)-y^(3))=n ["whene" n =(m^(3)+1)/(m^(3)-1)ne0]`
` x^(3)+y^(3) =n(x^(3)-y^(3)) and n ne0="variation constant."`
`(x^(3)+y^(3))prop(x^(3)-y^(3))(proved)`