If a power 1/3 + b power 1/3 + c power 1/3 then (a+b+c)whole cube = 27abc how?
Please cross check the given statement should be:
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.