Question

If x+y +z=0 then prove that x 3 +y 3 +z 3 =3xyz

Answer 1

Given, x3 + y3 + z3 = 3xyz

Therefore, x3 + y3 + z3 - 3xyz =0

This means,
x3 + y3 + z3 - 3xyz = (x + y + z) (x2 + y2 + z2 - xy - yz - zx)

Now if x + y + z = 0, then......

x3 + y3 + z3 - 3xyz = ( 0 ) (x2 + y2 + z2 - xy - yz - zx)  . . . . . . .... ..... ... .. . . .. . .....    [ subsituting the value of x + y + z ]

x3 + y3 + z3 - 3xyz = 0   

x3 + y3 + z3 = 3xyz .  

Answer 2

x+y+z=0 --------1   x³+y³+z³-3xyz=(x+y+z)(x²+y²+z²-xy-yz+zx)  --------2   Applying equation 1 in equation 2 x³+y³+z³-3xyz=0 ( x²+y²+z²-xy-yz-zx) x³+y³+z³-3xyz=0 x³+y³+z³=3xyz