Question

If x+y+z=0 prove that x(cube)+y(cube)+z(cube)=3xyz

Answer 1

w.k.t
(x+y+z)(x (s)+y (s)+z (s)-xy-yz-xz)=x (c)+y (c)+z (c)-3xyz substitute the value ok
(s)=square
(c)=cube

Answer 2

given x+y+z=0
to prove x³+y³+z³=3xyz
we have an identity 
(x+y+z )(x²+y²+z²-xy-yz-zx)=x³+y³+z³-3xyz
substituting x+y+z=0in the above equation,we get
0*(x²+y²+z²-xy-yz-zx)=x³+y³+z³-xy-yz-zx)
⇒x³+y³+z³-3xyz=0
⇒x³+y³+z³=3xyz