Question

If `x/a = y/b = z/c` prove that `(2x^3 - 3y^3 + 4z^3)/(2a^3 - 3b^3 + 4c^3) = ((2x - 3y + 4z)/(2a - 3b + 4c))^3`

Answer 1

Let `x/a = y/b = z/c = k`

Then x = ak, y = bk and z = ck

L.H.S = `(2x^3 - 3y + 4z)/(2a^3 - 3b^3 + 4c^3)`

`= (2(ak)^3 - 3(bk)^3 + 4(ck)^3)/(2a^3 - 3b^3 + 4c^3)`

`= (k^3(2a^3 - 3b^3 + 4c^2))/(2a^3 - 3b^3 + 4c^3)`

= `k^3`

RHS = `((2x - 3y + 4z)/(2a - 3b + 4c))^3`

`= ((2ak - 3bk + 4ck)/(2a - 3b + 4c))^3`

`= [(k(2a - 3b + 4c))/(2a - 3b + 4c)]^3`

`= k^3`

Hence LHS = RHS