Solve the following system of equations:

$$
\begin{aligned}
& x^{3}+y^{3}+z^{3}=8 \\
& x^{2}+y^{2}+z^{2}=22 \\
& \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=-\frac{z}{x y}
\end{aligned}
$$
According to equation (3), none of $x, y$, and $z$ can be 0. Bringing equation (3) to a common denominator and rearranging:

$$
\frac{x y+x z+y z+z^{2}}{x y z}=0
$$

The numerator of the fraction can be factored as:

$$
x y+x z+y z+z^{2}=(x+z)(y+z)
$$

Since the value of the product is 0, one of its factors must be 0. If $x=-z$, then from (1) we get $y^{3}=8$, so $y=2$. From (2) we then get $x_{2}=9$, which means either $x=3$ and $z=-3$, or $x=-3$ and $z=3$. In the case of $y=-z$, similarly, we find that $x=2$, and either $y=3$ and $z=-3$, or $y=-3$ and $z=3$.

Since none of the values obtained for $x, y$, and $z$ are 0, all our steps are reversible. Thus, the system of equations is satisfied by the following four triples:
$x_{1}=3$
$x_{2}=-3$
$x_{3}=2$
$x_{4}=2$
$y_{1}=2$
$y_{2}=2$
$y_{3}=3$
$y_{4}=-3$
$z_{1}=-3 ;$
$z_{2}=3$;
$z_{3}=-3 ;$
$z_{4}=3$.