Solve the following system of equations:

$$
\begin{aligned}
x^{3}+y^{3}+z^{3} & =8 \\
x^{2}+y^{2}+z^{2} & =22 \\
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{z}{x y} & =0
\end{aligned}
$$
Solution. From equation (3), it follows that none of $x, y, z$ can be 0. Multiplying this equation by $x y z$ and rearranging, we get:

$$
\begin{aligned}
y z + x z + x y + z^{2} & = 0 \\
z(x + z) + y(x + z) & = 0 \\
(z + x)(z + y) & = 0
\end{aligned}
$$

A product is zero if and only if at least one of its factors is zero.

Case I: $z = -x$. Substituting this into (1), we get $y^{3} = 8$, so $y = 2$. Finally, substituting the value of $y$ and $z = -x$ into (2), we get $x^{2} = 9$. From this, it follows that the triples that satisfy the conditions are those in which one of $x$ and $z$ is 3 and the other is -3:

$$
x_{1} = 3, \quad y_{1} = 2, \quad z_{1} = -3 ; \quad x_{2} = -3, \quad y_{2} = 2, \quad z_{2} = 3
$$

Case II: $z = -y$. Since the original system of equations is symmetric in $x$ and $y$, in this case we get the following two solutions:

$$
x_{3} = 2, \quad y_{3} = 3, \quad z_{3} = -3 ; \quad x_{4} = 2, \quad y_{4} = -3, \quad z_{4} = 3
$$

In neither case is any variable 0, so all four solutions satisfy the system of equations.