Problem 3. Determine all real values of $(x, y, z)$ for which

$$
\begin{array}{cccc}
x+y+z & = & 1 \\
x^{2} y+y^{2} z+z^{2} x & = & x y^{2}+y z^{2}+z x^{2} \\
x^{3}+y^{2}+z & = & y^{3}+z^{2}+x
\end{array}
$$
Solution. The second equation can be rewritten as

$$
(x-y)(y-z)(z-x)=0
$$

Now, since the third equation is not symmetric, we will distinguish 3 different cases:

(a) If $x=y$, the third equation becomes

$$
x^{2}+z=z^{2}+x
$$

or alternatively

$$
(x-z)(x+z-1)=0
$$

Therefore, from the last two, the options are $(\lambda, \lambda, \lambda)$ or $(\lambda, \lambda,-\lambda+1)$. Substituting into the first, we directly obtain two solutions of the system, which are $(1 / 3,1 / 3,1 / 3)$ and $(0,0,1)$.

(b) If $x=z$, the third equation becomes

$$
x^{3}+y^{2}=y^{3}+x^{2}
$$

or alternatively

$$
(x-y)\left(x^{2}+y^{2}+x y-x-y\right)=0
$$

From here, substituting into the first equation, we obtain two new solutions (in addition to the one corresponding to $x=y=z$), which are $(0,1,0)$ and $(2 / 3,-1 / 3,2 / 3)$.

(c) If $y=z$, the third equation becomes

$$
x^{3}+y=y^{3}+x
$$

or alternatively

$$
(x-y)\left(x^{2}+x y+y^{2}-1\right)=0
$$

From here, we obtain two new solutions, $(1,0,0)$ and $(-1,1,1)$

Therefore, the system has the six solutions we have found.