[ Systems of nonlinear algebraic equations ] Higher degree equations (miscellaneous) $\quad]$

Solve the system of equations:

$x^{3}-y=6$,

$y^{3}-z=6$,

$z^{3}-x=6$.
If $x \geq y$, then $y=x^{3}-6 \geq y^{3}-6=z$. Similarly, from this we get that $z \geq x$. Therefore, $x \geq y \geq z \geq x$, from which $x=y=z$.

If $x \leq y$, then similarly we prove that $x=y=z$.

Thus, all equations of the original system are equivalent to the equation $x^{3}-x=6$. One of its roots $-x=$ 2 is obvious. Dividing $x^{3}-x-6$ by $x-2$, we get $x^{2}+2 x+3$. But the equation $x^{2}+2 x+3=0$ has no roots.

## Answer

$(2,2,2)$.