Question

x+y+z=0 then prove x^3+y^3+z^3=3xyz

Answer 1

Suppose an example let keep the value of x be -2 ,y be +1 and z be +1
So according the first equation it comes to be -2+1+1 = 0

But in second one
-2^3 + 1^3 + 1^3 it comes to be -6

Solving RHS ... 3(-2)(1)(1) IT DO COMES -6 hence proved you can choose any value but it should follow the clinging criteria of equation 1 thou

Answer 2

We know that
a³+b³+c³-3abc=(a+b+c)(a²+b²+c²-ab-bc-ca)
Sox³+y³+z³-3xyz=(x+y+z)(x²+y²+z²-xy-yz-zx)
Givenx+y+z=0
Therefore
x³+y³+z³-3xyz=(0)(x²+y²+z²-xy-yz-zx)
x³+y³+z³-3xyz=o
x³+y³+z³=3xyz
Therefore , if x+y+z=0 , x³+y³+z³=3xyz