Example 6 Solve the system of equations: $\left\{\begin{array}{l}a+b+c=0, \\ a^{2}+b^{2}+c^{2}=1, \\ a^{3}+b^{3}+c^{3}=4 a b c .\end{array}\right.$
(2017, Irish Mathematical Olympiad)
【Analysis】The three equations are all related to the basic symmetric expressions of $a, b, c$, and we can find their connections from relevant formulas.
Solution Note that,
$$
\begin{array}{l}
a^{3}+b^{3}+c^{3}-3 a b c \\
=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right) \\
=0 .
\end{array}
$$

From $3 a b c=a^{3}+b^{3}+c^{3}=4 a b c \Rightarrow a b c=0$.
It is easy to see that among $a, b, c$, exactly one is 0.
Due to symmetry, assume $c=0$.
Then $a+b=0, a^{2}+b^{2}=1$.
Solving gives $(a, b, c)=\left( \pm \frac{\sqrt{2}}{2}, \mp \frac{\sqrt{2}}{2}, 0\right)$.
Therefore, the solutions to the original system of equations are
$$
\begin{aligned}
(a, b, c)= & \left( \pm \frac{\sqrt{2}}{2}, \mp \frac{\sqrt{2}}{2}, 0\right),\left( \pm \frac{\sqrt{2}}{2}, 0, \mp \frac{\sqrt{2}}{2}\right), \\
& \left(0, \pm \frac{\sqrt{2}}{2}, \mp \frac{\sqrt{2}}{2}\right) .
\end{aligned}
$$