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.
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 $(1 + m)^{5} + (1 + \omega m)^{5} + (1 + \omega^{2} m)^{5} = 3$. Expanding and noting that
\[ 1 + \omega^{k} + \omega^{2 k} = \begin{cases} 
0, & \text{if } k \text{ is not a multiple of 3,} \\
3, & \text{if } k \text{ is a multiple of 3,}
\end{cases} \]
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 $x^{3} + y^{3} + z^{3} - 3 x y z = (x + y + z) \cdot (x + y \omega^{2} + z \omega) \cdot (x + y \omega + z \omega^{2}) = 0$, we know that the equation $x^{3} - 3 y z \cdot x + (y^{3} + z^{3}) = 0$ has three roots $-y - z, -y \omega^{2} - z \omega, -y \omega - z \omega^{2}$, where $y, z$ are the roots of the quadratic equation $A^{2} - q A - \frac{p^{3}}{27} = 0$.