12.119 Solve the system of equations
$$
\left\{\begin{array}{l}
x+y+z=0, \\
x^{3}+y^{3}+z^{3}=-18 .
\end{array}\right.
$$

for integer solutions.
(China High School Mathematics League; 1978)
[Solution] From (1), we get
$$
z=-(x+y) .
$$

Substituting (3) into (2) yields $x^{3}+y^{3}-(x+y)^{3}=-18$, simplifying to
From (3)
$$
\begin{aligned}
x y(x+y) & =6 . \\
x y z & =-6 .
\end{aligned}
$$

From (4), we know that $x, y, z$ are divisors of 6, and they must satisfy equations (1) and (2). Therefore, among $x, y, z$, there is exactly one negative number, and the absolute value of this negative number should be the largest.
Let $x=-3$, then we get
$$
\left\{\begin{array} { l } 
{ y = 1 , } \\
{ z = 2 , }
\end{array} \text { or } \left\{\begin{array}{l}
y=2, \\
z=1 .
\end{array}\right.\right.
$$

Similarly, by letting $y=-3, z=-3$, we can find the other four solutions.
Thus, the system of equations has six solutions, which are
$$
\begin{array}{l}
\left\{\begin{array} { l } 
{ x = - 3 , } \\
{ y = 1 , } \\
{ z = 2 , }
\end{array} \quad \left\{\begin{array}{l}
x=-3, \\
y=2, \\
z=1,
\end{array},\left\{\begin{array}{l}
x=1, \\
y=-3, \\
z=2,
\end{array}\right.\right.\right. \\
\left\{\begin{array}{l}
x=2, \\
y=-3, \\
z=1,
\end{array},\left\{\begin{array}{l}
x=1, \\
y=2, \\
z=-3,
\end{array},\left\{\begin{array}{l}
x=2, \\
y=1, \\
z=-3 .
\end{array}\right.\right.\right.
\end{array}
$$