Example 2 Solve the system of simultaneous equations
$$
\left\{\begin{array}{l}
x+y+z=3, \\
x^{2}+y^{2}+z^{2}=3, \\
x^{5}+y^{5}+z^{5}=3 .
\end{array}\right.
$$

Find all real or complex roots.
(2nd USA Mathematical Olympiad)
Let $T_{n}=x^{n}+y^{n}+z^{n} (n=1,2, \cdots)$, then $T_{1}=T_{2}=T_{5}=3$.
Let $x, y, z$ be the three roots of the cubic polynomial
$$
f(t)=(t-x)(t-y)(t-z)=t^{3}-b_{1} t^{2}+b_{2} t-b_{3}
$$

Then, by the given conditions, we have $b_{1}=T_{1}=3, b_{2}=\frac{1}{2}\left(T_{1}^{2}-T_{2}\right)=3$. Next, we calculate $b_{3}$.
By Newton's formulas, we have
$$
\begin{array}{l}
T_{5}=3 T_{4}-3 T_{3}+b_{3} T_{2}, \\
T_{4}=3 T_{3}-3 T_{2}+b_{3} T_{1}, \\
T_{3}=3 T_{2}-2 T_{1}+b_{3} \cdot 3 .
\end{array}
$$

From the above three equations, we get $T_{5}=30 b_{3}-27$.
Since $T_{5}=3$, we obtain $b_{3}=1$.
Thus, $f(t)=t^{3}-3 t^{2}+3 t-1=(t-1)^{3}$.
Therefore, $x=y=z=t=1$ is the solution.