5. (7) Solve the system of equations $x+y=y^{2} z$, $y+z=z^{2} x$, $z+x=x^{2} y$.
5. Note that if $x, y, z$ is a solution, then $-x, -y, -z$ is also a solution. Also note that if one of the variables is 0, then all the others are also 0. From now on, we will assume that $x, y, z \neq 0$.

Suppose that $x, y$, and $z$ are not all of the same sign. Then, without loss of generality, we can assume that $x>0, y>0$, and $z<0$. Then $x+y>y+z$, but $y^{2} z<y>z$, or $xy>z$. Then $y^{2} z<z+x$. Contradiction. Similarly, if $x<0$, then $x+y<z+x$, but $x^{2} y>y^{2} z$.

Therefore, all variables are equal. We obtain three solutions: $x=y=z=0, x=y=z= \pm \sqrt{2}$.