2. Solve the system of equations $\left\{\begin{array}{l}x+y+z=3 \\ x^{2}+y^{2}+z^{2}=3 \\ x^{5}+y^{5}+z^{5}=3\end{array}\right.$.
2. Let $s_{n}=x^{n}+y^{n}+z^{n}(n=1,2, \cdots)$. Then $s_{1}=s_{2}=s_{5}=3$. Suppose $x, y, z$ are the three roots of the cubic polynomial $f(t)=(t-x)(t-y)(t-z)=t^{3}-\sigma_{1} t^{2}+\sigma_{2} t-\sigma_{3}$, then by the given conditions $\sigma_{1}=s_{1}=3, \sigma_{2}=\frac{1}{2}\left(s_{1}^{2}-s_{2}\right)=3$. Next, we calculate $\sigma_{3}$. By Newton's formulas, we have
$$
\begin{array}{l}
s_{5}=3 s_{4}-3 s_{3}+\sigma_{3} s_{2}, \\
s_{4}=3 s_{3}-3 s_{2}+\sigma_{3} s_{1}, \\
s_{3}=3 s_{2}-2 s_{1}+3 \sigma_{3} .
\end{array}
$$

From the above three equations, we get $s_{5}=30 \sigma_{3}-27$. Since $s_{5}=3$, then $\sigma_{3}=1$. Therefore, $f(t)=t^{3}-3 t^{2}+3 t-1=(t-1)^{3}$, hence $x=$ $y=z=1$.