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

According to Dickson[1], S. Realis gave the following parametric solution.
(x,y,z) = ( (a-b)^3+(a-c)^3,  (b-a)^3+(b-c)^3,  (c-a)^3+(c-b)^3 ).

We show x^3 + y^3 + z^3 = 3xyz has a simple parametric solution.


x^3 + y^3 + z^3 = 3xyz.......................................................(1)

We use famous identity below.

x^3 + y^3 + z^3 - 3xyz = (x+y+z)(x^2+y^2+z^2-yz-xz-yx).

x^2+y^2+z^2-yz-xz-yx is positive except the case x=y=z.

Hence x+y+z must be 0 if we exclude the case x=y=z.

Let x=a, y=b, z=-a-b then we obtain a simple(trivial?) parametric solution.

(x,y,z)=(a,b,-a-b).

a,b are arbitrary.

Realis's solution satisfies the condition x+y+z =0.

Reference

[1]: Dickson: History of the theory of numbers,  vol II