Question

If `x^(1/3) + y^(1/3) + z^(1/3) = 0`, prove that (x + y + z)³ = 27xyz

Answer 1

`x^(1/3) + y^(1/3) + z^(1/3)` = 0

⇒ `(x^(1/3) + y^(1/3)) + z^(1/3)` = 0 cubing both sides, we get :

⇒ `(x^(1/3) + y^(1/3))^3 + z + 3 (x^(1/3) + y^(1/3)) z^(1/3) (x^(1/3) + y^(1/3) + z^(1/3))` = 0

⇒ `x + y+ 3 x^(1/3)y^(1/3) (x^(1/3) + y^(1/3)) + z + 0` = 0

⇒ `x + y+ 3 x^(1/3)y^(1/3)(-z^(1/3)) + z` = 0    ...(Using the given condition again)

⇒ x + y + z = `3 x^(1/3)y^(1/3)z^(1/3)`
⇒ (x + y + z)³ = 27xyz.