Solve the following system of equations:

$$
\begin{aligned}
& x^{2}+2 y z=x \\
& y^{2}+2 z x=y \\
& z^{2}+2 x y=z
\end{aligned}
$$
From the sum of the three equations

$$
(x+y+z)^{2}=x+y+z
$$

thus $x+y+z=1$, or $x+y+z=0$. From the difference of the first two equations

$$
(x-y)(x+y-2 z-1)=(x-y)[(x+y+z)-(3 z+1)]=0
$$

and similarly from the difference of the second and third equations

$$
(y-z)[(x+y+z)-(3 x+1)]=0
$$

If $x+y+z=1$, then (1) and (2) transform as follows:

$$
\begin{gathered}
(x-y) 3 z=0 \\
(y-z) \cdot 3 x=0
\end{gathered}
$$

From this, $x y=y z=z x$ follows, so if none of $x, y$, and $z$ are zero, they are equal, and since their sum is 1, $x=y=z=\frac{1}{3}$. However, if one of them is zero, then exactly two of them are zero, and thus the third is 1. This gives us three more solutions corresponding to $\{x, y, z\}=\{0 ; 0 ; 1\}$.

If $x+y+z=0$, then let us introduce the unknowns

$$
x^{\prime}=x+\frac{1}{3}, \quad y^{\prime}=y+\frac{1}{3}, \quad z^{\prime}=z+\frac{1}{3}
$$

For these, on the one hand, $x^{\prime}+y^{\prime}+z^{\prime}=1$, and on the other hand, (1) and (2) now take the form

$$
\begin{aligned}
& \left(x^{\prime}-y^{\prime}\right) \cdot 3 x^{\prime}=0 \\
& \left(y^{\prime}-z^{\prime}\right) \cdot 3 x^{\prime}=0
\end{aligned}
$$

which is identical to the system of equations obtained in the first case. From the solution $x^{\prime}=y^{\prime}=z^{\prime}=\frac{1}{3}$, we now get $x=y=z=0$, and from the other three solutions, we get the solutions corresponding to $\{x, y, z\}=\left\{-\frac{1}{3},-\frac{1}{3} ; \frac{2}{3}\right\}$.

Therefore, the system of equations has eight solutions; the following triples:

$$
(0 ; 0 ; 0),\left(\frac{1}{3} ; \frac{1}{3} ; \frac{1}{3}\right),(1 ; 0 ; 0),(0 ; 1 ; 0),(0 ; 0 ; 1),\left(\frac{2}{3} ;-\frac{1}{3} ;-\frac{1}{3}\right),\left(-\frac{1}{3} ; \frac{2}{3} ;-\frac{1}{3}\right),\left(-\frac{1}{3} ;-\frac{1}{3} ; \frac{2}{3}\right)
$$