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

[Edwin Clark]

On Mon, 24 Apr 2006, Emeric Deutsch wrote:

> Dear Seqfans,
> Do you know any literature on the solutions of
> x^3 + y^3 + z^3 = 3xyz in integers?
> Any other solutions than x=y=z=1,2,3,... ?

Since 

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. For example things like,

n, -k, -(n-k)

PS I found the above factorization at 

   http://www.thiel.edu/mathproject/atps/chptr08/p072.htm

and verified it using Maple.