$4 \cdot 100$ Solve the system of equations $\quad\left\{\begin{array}{l}x+y+z=0, \\ x^{3}+y^{3}+z^{3}=-18 .\end{array}\right.$ for integer solutions.
[Solution] From (1), we get $z=-(x+y)$
Substituting into (2) yields $x^{3}+y^{3}-(x+y)^{3}=-18$,
Simplifying, we get $xy(x+y)=6$.
From (3), we get $xyz=-6$,
Therefore, $x, y, z$ must be divisors of 6.
From (1) and (2), we know that among $x, y, z$, there is exactly one negative number, and the absolute value of this negative number should be greater than the absolute values of the other two numbers. Therefore, this negative number must be -3, and the other two numbers are 1 and 2. Upon verification, when $x, y, z$ are $-3, 1, 2$ respectively, equations (1) and (2) both hold. By symmetry, the given system of equations has six solutions:
$$
\begin{array}{l}
\left\{\begin{array} { l } 
{ x = - 3 , } \\
{ y = 1 , } \\
{ z = 2 ; }
\end{array} \quad \left\{\begin{array} { l } 
{ x = - 3 , } \\
{ y = 2 , } \\
{ z = 1 ; }
\end{array} \quad \left\{\begin{array}{l}
x=1, \\
y=-3, \\
z=2 ;
\end{array}\right.\right.\right. \\
\left\{\begin{array} { l } 
{ x = 2 , } \\
{ y = - 3 , } \\
{ z = 1 . }
\end{array} \quad \left\{\begin{array} { l } 
{ x = 1 , } \\
{ y = 2 } \\
{ z = - 3 ; }
\end{array} \quad \left\{\begin{array}{l}
x=2 \\
y=1 \\
z=-3
\end{array}\right.\right.\right. \\
\end{array}
$$