So $x^3-x=y^3-y=z^3-z=s$ for some number $s$. If two variables are equal, say $x=y$, then we get $x^3+x+z=-x^2z$ and hence $(x^2+1)(x+z)=0$ so that $z=-x$ and hence $s=0$ so that $x^3=x$ and hence we get the solutions $(0,0,0), (1,1,-1), (1,-1,1), (-1,1,1), (-1,-1,1), (-1,1,-1), (1,-1,-1)$. On the other hand, if all variables are distinct, they are the three solutions of $T^3-T-s=0$ and hence by Vieta, we get $x+y+z=0$ and $xyz=s$ so that the equations become $-s=x^3+y+z=s+x+y+z=s$ and hence $s=0$ so that $(x,y,z)$ is any of the six permutations of $(0,1,-1)$.

interesting