2017/003) Show that for a,b,c all different $\sqrt[3]{a-b} + \sqrt[3]{b-c} + \sqrt[3]{c-a} != 0$

if  $\sqrt[3]{a-b} + \sqrt[3]{b-c } + \sqrt[3]{c-a} = 0$
then using $(x + y + z = 0) =>x^3 + y^3 + z^3 = 3xyz$
we get $0 = 3\sqrt[3]{(a-b)(b-c)(c-a)}$ giving a = b or b= c or c=a.
so if all are different result cannot be zero.