If a power 1/3 + b power 1/3 + c power 1/3 then (a+b+c)whole cube = 27abc how?

a power 1/3 + b power 1/3 + c power 1/3 = 0

Now, let a power 1/3 = x, b power 1/3 = y, c power 1/3 = z

So, x+y+z=0

Using the identity, we get,

x power 3 + y power 3 + z power 3 = 3xyz

This gives,

(a power 1/3) power 3 + (b power 1/3) power 3 + (c power 1/3) power 3 = 3(a power 1/3) (b power 1/3) (c power 1/3)

a + b + c = 3(a power 1/3) (b power 1/3) (c power 1/3)

Cubing both sides, we get,

(a + b + c)^{3} = 27abc

Hence, proved.