Solve the system of equations:

$$
\begin{aligned}
& x^{3}-y=6 \\
& y^{3}-z=6 \\
& z^{3}-x=6
\end{aligned}
$$
Let the smallest of the numbers $x, y, z$ be, for example, $x$. By comparing the first and third equations, we then get that $x^{3}-y=z^{3}-x$, which means

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

According to our assumption, on the one hand $x \leq y$, and on the other hand $x \leq z$, so $x^{3} \leq z^{3}$. The left side of (1) can only be equal to the right side if $x=y$ and $x^{3}=z^{3}$, thus $x=z$.

For every real solution of the system of equations, therefore, $x=y=z$. Hence, $x^{3}-x=6$, that is,

$$
0=x^{3}-x-6=(x-2)\left(x^{2}+2 x+3\right)
$$

Since $x^{2}+2 x+3=(x+1)^{2}+2>0$, the only real solution of the system of equations is

$$
x=y=z=2
$$