Question

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

Answer 1

x+y+z=0 --------(1)

lhs = x^3+y^3+z^3
 = (x+y+z)(x^2+y^2+z^2 -xy-yz-zx)+3xyz  [ identity]
= 0(x^2+y^2+z^2 -xy-yz-zx) +3xyz [from (1)]

= 3xyz 
=rhs

Answer 2

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 .