I'm interested in the system of equations:
$a(b^2+c)=c(c+ab)$
$b(c^2+a)=a(a+bc)$
$c(a^2+b)=b(b+ac)$
It is easy to see that $a=b=c=t$ are solutions for all $t$, in fact these are the only real solutions.
If we allow complex solutions there are in fact 6 more solutions, after a lot of elementary algebra you can show that the solutions for a are the solutions $(a^3+(-1-i)a^2+a-1)(a^3+(-1+i)a^2+a-1)=0$ and from this you can recover b and c. Can someone give a simple argument for this fact? The argument I have is straightforward but a bit too involved and I have a feeling there might be something easy here.