Three. (Total 25 points) Find all integer solutions to the system of equations $\left\{\begin{array}{l}x+y+z=3, \\ x^{3}+y^{3}+z^{3}=3\end{array}\right.$.
$$
\begin{array}{l} 
\text { 3. } x+y=3-z, \\
x^{3}+y^{3}=3-z^{3} .
\end{array}
$$
(1) - (2) gives $xy=\frac{8-9z+3z^{2}}{3-z}$.

Knowing that $x, y$ are the roots of $t^{2}-(3-z)t+\frac{8-9z+3z^{2}}{3-z}=0$.
Solving gives $t_{1,2}=\frac{(3-z) \pm \sqrt{(3-z)^{2}-4 \times \frac{8-9z+3z^{2}}{3-z}}}{2}$.
$$
\begin{aligned}
\text { Also, } & (3-z)^{2}-4 \times \frac{8-9z+3z^{2}}{3-z}=(z-1)^{2} \cdot \frac{z+5}{z-3} \\
& =(z-1)^{2} \cdot\left(1+\frac{8}{z-3}\right) .
\end{aligned}
$$

When $z-1=0$ i.e., $z=1$, $x=y=1$;
When $z-3=1$ i.e., $z=4$, $x=4$ or $-5$, $y=-5$ or $4$;
When $z-3=-8$ i.e., $z=-5$, $x=y=4$.
Thus, there are four solutions: $(1,1,1),(4,-5,4),(-5,4,4)$, $(4,4,-5)$.