10.1. Find all triples of real numbers $(x, y, z)$ that are solutions to the system of equations $\left\{\begin{array}{l}x^{2}+y+z=1, \\ x+y^{2}+z=1, \\ x+y+z^{2}=1 .\end{array}\right.$
Answer. $(1,0,0),(0,1,0),(0,0,1),(-1-\sqrt{2},-1-\sqrt{2},-1-\sqrt{2}),(-1+\sqrt{2},-1+\sqrt{2},-1+\sqrt{2})$.

Solution. By factoring the difference of the first and second equations, we get $(x-y)(x+y-1)=0$, which means that for the pair of variables $x, y$ from the triple of solutions of the system $(x, y, z)$, one of the two equalities $x=y$ or $x+y=1$ is satisfied. The same is true for the pairs of variables $y, z$ and $x, z$. If the second equality is satisfied for at least one of these pairs, for example, $x+y=1$, then from the third equation $z=0$ and from the first equation $x^{2}+1-x=1 \Leftrightarrow x^{2}-x=0$, from which either $x=1, y=z=0$, or $x=0, y=1, z=0$. Combining such triples for all pairs of variables, we get three solutions of the system, in each of which one variable is 1 and the others are 0.

If for each pair of variables from the solution of the system the first equation is satisfied, then all variables are equal to each other and equal to any of the roots of the equation $x^{2}+2 x-1=0$, that is, to one of the numbers $-1+\sqrt{2},-1-\sqrt{2}$, which gives two more solutions of the system.

Grading criteria. Answers: 2 points. Proof that there are no other answers: 5 points.