Find all integer solutions of the following system of equations:

$$
x+y+z=3 ; \quad x^{3}+y^{3}+z^{3}=3 .
$$
The following identity is known:

$$
(x+y+z)^{3}-\left(x^{3}+y^{3}+z^{3}\right)=3(x+y)(x+z)(y+z)
$$

Given that \(x+y+z=x^{3}+y^{3}+z^{3}=3\), we have:

$$
\begin{gathered}
3(x+y)(x+z)(y+z)=3^{3}-3=24, \quad \text { hence } \\
(x+y)(x+z)(y+z)=8
\end{gathered}
$$

The factors on the left-hand side are integers, and since \(x+y+z=3\), the sum of the factors is 6, which is even. Therefore, an even number of the factors, either 0 or 2, must be odd.

If all factors are even, then each of their absolute values is 2, and since their sum is 6, we have:

$$
x+y=x+z=y+z=2
$$

Then \(x=x+y+z-(y+z)=3-2=1\), and similarly \(y=z=1\).

If two factors are odd, then their values are +1 or -1, so the third factor is +8 or -8. The latter is not possible, because in that case the sum of the factors could be at most -6. Therefore, the only even factor is 8, and the other two are -1.

Three cases are possible, one of which is:

$$
x+y=8 \quad x+z=y+z=-1
$$

Then \(x=x+y+z-(y+z)=3+1=4\), and similarly \(y=4, z=-5\). By appropriately swapping the factors, we get two more solutions.

Thus, the equation is satisfied by four triples of integers, the correctness of which can be easily verified by substitution:

$$
x=y=z=1 ; \quad x=y=4, \quad z=-5 ; \quad x=z=4, \quad y=-5 ; \quad y=z=4, \quad x=-5
$$