Solve the following system of equations

$$
x y+x z=8-x^{2} ; \quad x y+y z=12-y^{2} ; \quad y z+z x=-4-z^{2}
$$
In all three equations, if we collect the unknowns on the left side, then on the left side of the sum of the equations, a complete square stands:

$$
\begin{gathered}
x(x+y+z)=8 \\
y(x+y+z)=12 \\
z(x+y+z)=-4
\end{gathered}
$$

From the sum of the three equations,

$$
(x+y+z)^{2}=16
$$

From this, $x+y+z=4$, or $x+y+z=-4$. Substituting the values obtained for $(x+y+z)$ on the left side of (1), (2), (3), we get the solutions: If $x+y+z=4$, then $x=2, y=3, z=-1$, and if $x+y+z=-4$, then $x=-2$, $y=-3, z=1$.