8 Utilize the discriminant
Example 10. Try to determine all real solutions of the system of equations
$$
\left\{\begin{array}{l}
x+y+z=3, \\
x^{2}+y^{2}+z^{2}=3, \\
x^{5}+y^{5}+z^{5}=3
\end{array}\right.
$$
Solving: Since $x^{2}+y^{2}=(x+y)^{2}-2xy=(3-z)^{2}-2xy=3-z^{2}$, we have, from the first two equations,
$$
x+y=3-z, \quad xy=z^{2}-3z+3.
$$

From this, we know that $x, y$ are the two real roots of the quadratic equation in $t$
$$
t^{2}-(3-z)t+\left(z^{2}-3z+3\right)=0.
$$

By the discriminant theorem, we get
$$
\begin{aligned}
\Delta & =(z-3)^{2}-4\left(z^{2}-3z+3\right) \\
& =-3(z-1)^{2} \geqslant 0.
\end{aligned}
$$

Solving this inequality yields $z=1$.
When $z=1$, it is easy to get $x=y=1$. Substituting into (3), it fits, so the real solution of the original system of equations is
$$
x=1, \quad y=1, \quad z=1.
$$