Example 5 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 $x, y, z$ be the roots of the cubic equation $r^{3}-a r^{2}+b r-c=0$, then $a=x+y+z=3, b=$ $xy+yz+zx, c=xyz$. By squaring both sides of $x+y+z=3$ and rearranging, we get $b=3$, thus the cubic equation becomes $r^{3}-3 r^{2}+3 r-c=0$.

Let $c=1+m^{3}$, then $(r-1)^{3}=m^{3}$, so $r=1+m, 1+\omega m, 1+\omega^{2} m\left(\omega=-\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}\right)$. From (3), we have $(1+m)^{5}+(1+\omega m)^{5}+\left(1+\omega^{2} m\right)^{5}=3$. Expanding and noting that
\[ 1+\omega^{k}+\omega^{2 k}=\left\{\begin{array}{l}0, \text{ if } k \text{ is not a multiple of } 3, \\ 3, \text{ if } k \text{ is a multiple of } 3,\end{array}\right. \]
we get $30 m^{3}=0$ or $m=0$, thus all the roots are $x=y=z=1$.
Here, we also point out: using $\omega$, we can derive the solution formula for the real-coefficient cubic equation of the form $x^{3}+p x+q=0$ (for the form $y^{3}+a y^{2}+b y+c=0$, it can be transformed to this form by the substitution $y=x-\frac{a}{3}$).

In fact, from $x^{3}+y^{3}+z^{3}-3 x y z=(x+y+z) \cdot\left(x+y \omega^{2}+z \omega\right) \cdot\left(x+y \omega+z \omega^{2}\right)=$ 0, we know that the equation $x^{3}-3 y z \cdot x+\left(y^{3}+z^{3}\right)=0$ has three roots $-y-z,-y \omega^{2}-z w,-y \omega-z \omega^{2}$, where $y, z$ are the roots of the quadratic equation $A^{2}-q A-\frac{p^{3}}{27}=0$.