Find all ordered triples $(x, y, z)$ of real numbers satisfying the following system of equations:

$$
\begin{aligned}
x^{3} & =\frac{z}{y}-2 \frac{y}{z} \\
y^{3} & =\frac{x}{z}-2 \frac{z}{x} \\
z^{3} & =\frac{y}{x}-2 \frac{x}{y}
\end{aligned}
$$
We have

$$
\begin{aligned}
& x^{3} y z=z^{2}-2 y^{2} \\
& y^{3} z x=x^{2}-2 z^{2} \\
& z^{3} x y=y^{2}-2 x^{2}
\end{aligned}
$$

with $x y z \neq 0$.

Adding these up we obtain $\left(x^{2}+y^{2}+z^{2}\right)(x y z+1)=0$. Hence $x y z=-1$. Now the system of equations becomes:

$$
\begin{aligned}
& x^{2}=2 y^{2}-z^{2} \\
& y^{2}=2 z^{2}-x^{2} \\
& z^{2}=2 x^{2}-y^{2}
\end{aligned}
$$

Then the first two equations give $x^{2}=y^{2}=z^{2}$. As $x y z=-1$, we conclude that $(x, y, z)=$ $(1,1,-1),(1,-1,1),(-1,1,1)$ and $(-1,-1,-1)$ are the only solutions.