If the triplicate of the ratio `(x+a) : (x + b) " is " a : b`, then prove that `x^(3) - 3abx - ab(a +b) = 0`.

Explanation

The triplicate of `(x +a) : (x + b) " is " ((x+a)^(3))/((x+b)^(3))`
As per question, ` ((x+a)^(3))/((x+b)^(3)) = a/b`.
or, `a (x + b)^(3) = b (x + a)^(3)`
or, `a (x^(3) + 3x^(2)b + 3xb^(2) + b^(3)) = b (x^(3) + 3x^(2)a + 3xa^(2)+a^(3))`
or, `ax^(3)+ 3abx^(2) + 3ab^(2) x + ab^(3) = bx^(3) + 3abx^(2) + 3a^(2) bx + a^(3)b`.
or, `ax^(3) - bx^(3) + 3ab^(2) x - 3a^(2) bx + ab^(3) - a^(3) b = 0 `.
or, `x^(3) (a -b) - 3abx (a -b) - ab (a^(2)-b^(2)) = 0`
or, `x^(3) (a-b) - 3abx (a -b) - ab (a +b) (a-b) = 0`
or, `(a-b){x^(3) - 3abx - ab(a+b)} = 0`
or, `x^(3) - 3abx - ab (a + b) = 0 " "[:' a ne b , :. a -b ne 0]`
` :. x^(3) - 3abx - ab (a +b) = 0`.