18.2.7 ** Find all integer solutions of the system of equations $\left\{\begin{array}{l}x^{3}+y^{3}+z^{3}=3 \\ x+y+z=3\end{array}\right.$ .
According to the identity
$$
(x+y+z)^{3}-\left(x^{3}+y^{3}+z^{3}\right)=3(x+y)(x+z)(y+z),
$$

if integers $x, y, z$ satisfy the original equation, then
$$
8=(x+y)(y+z)(x+z)=(3-x)(3-y)(3-z) .
$$

Since $x+y+z=3$, we have
$$
6=(3-x)+(3-y)+(3-z) .
$$

From equation (3), we know that $3-x, 3-y, 3-z$ are either all even or only one is even. In the first case, according to equation (2), the absolute values of these three numbers are all equal to 2, and thus, by equation (3), they are all equal to 2. Therefore, $x=y=z=1$. In the second case, according to equation (2), one of $3-x, 3-y, 3-z$ has an absolute value of 8, and the other two have an absolute value of 1. By equation (3), one of these numbers is 8, and the other two are -1. Thus, $x=-5, y=z=4$, or $x=y=4, z=-5$, or $x=4, y=-5, z=4$.
Therefore, the system of equations has only four integer solutions, namely $x, y, z$ are $1,1,1 ;-5,4,4 ; 4,-5$,
$$
4 ; 4,4,-5 .
$$