x^3 + y^3 + z^3 = 3xyz

[Edwin Clark]

> 
(*)  x^3+y^3+z^3-3*x*y*z = 
    (x-z-z*alpha+y*alpha)*(x-y-y*alpha+z*alpha)*(z+x+y)
> 
> where alpha is a root of x^2 + x + 1,
> 
> any roots of x+y+z=0 will be a solution. ...
> ___________________________________________________


The only integer solutions are x=y=z and z = -x-y. 
Suppose that x,y,z is another solution and x+y+z is not 0
then by the above factorization one of the other factors
must be 0. Say  x-z-z*alpha+y*alpha = 0. Since alpha is
not a rational number, y-z = 0, that is, y = z and hence
also x=z. So x=y=z. If the other factor involving alpha
is 0 a similar argument shows x = y = z.