10.1. Solve the system of equations in real numbers: $x^{3}+y=2 \mathrm{x}, y^{3}+x=2 \mathrm{y}$.
Answer. $(0,0), \pm(1,1), \pm(\sqrt{3},-\sqrt{3})$ - a total of 5 solutions.

Solution. Consider the cases.

1) $x=y$, then $x^{3}=x=y$, from which $(x, y)=(0,0), \pm(1,1)$ - the first three solutions.
2) $x=-y$, then $x^{3}=3 x=-y$, from which $(x, y)=(0,0), \pm(\sqrt{3},-\sqrt{3})$ - the first and the last two solutions. (Joke - no need to solve further)
3) $x \neq \pm y$ and $x, y \neq 0$. Subtract the second equation from the first and divide by $x-y$, we get $x^{2}+x y+y^{2}=3$. Add the first and second equations and divide by $x+y$, we get $x^{2}-x y+y^{2}=1$. Therefore, $x^{2}+y^{2}=2, x+y=1$, from which $x^{2}+y^{2}-2 x y=0$, that is, $x=y$, which is impossible in this case. No new solutions are obtained.

Grading criteria. All solutions are guessed and verified: 1 point. The first two cases are considered: 1 point for each. Solutions are lost: deduct 1 to 4 points.