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

[Emeric Deutsch]

Thanks to all of you!!
It seems that the simple-looking identity
 	x^3+y^3+z^3-3xyz = (x+y+z)*[(x-y)^2+(x-z)^2+(y-z)^2]/2 
gives beautifully the integer solutions of x^3+y^3+z^3=3xyz.

Emeric Deutsch