Example 2 Find all triples $(a, b, c)$ such that $a b+b c+c a=1$, and $a^{2} b+c=b^{2} c+a=c^{2} a+b$ hold.
Solve:
$$
\begin{array}{l}
a^{2} b+c=b^{2} c+a \\
\Rightarrow a(1-b c-c a)+c=b(1-a b-c a)+a \\
\Rightarrow a b^{2}+c=c a^{2}+b .
\end{array}
$$

Similarly, $b c^{2}+a=a b^{2}+c$.
Then $a b^{2}+c=b c^{2}+a=c a^{2}+b$
$$
\begin{array}{l}
\Rightarrow a^{2} b+c-\left(a b^{2}+c\right)=b^{2} c+a-\left(b c^{2}+a\right) \\
\Rightarrow b\left(a^{2}+c^{2}-a b-b c\right)=0 .
\end{array}
$$

Similarly,
$$
\begin{array}{l}
c\left(b^{2}+a^{2}-b c-c a\right)=0, \\
a\left(c^{2}+b^{2}-c a-a b\right)=0 .
\end{array}
$$

If $a=0$, then substituting into the two sets of equations in the problem, we get
$$
b c=1 \text {, and } b=c
$$
$\Rightarrow b=c=1$ or $b=c=-1$
$$
\Rightarrow(a, b, c)=(0, \pm 1, \pm 1) \text {. }
$$

Similarly, when $b=0$ or $c=0$, we get
$$
(a, b, c)=( \pm 1,0, \pm 1),( \pm 1, \pm 1,0) \text {. }
$$

If $a b c \neq 0$, then
$$
\begin{array}{l}
a^{2}+c^{2}-a b-b c=0, \\
b^{2}+a^{2}-b c-c a=0, \\
c^{2}+b^{2}-c a-a b=0 .
\end{array}
$$

Adding the three equations and simplifying, we get
$$
\begin{array}{l}
(a-b)^{2}+(b-c)^{2}+(c-a)^{2}=0 \\
\Rightarrow a=b=c .
\end{array}
$$

Combining with $a b+b c+c a=1$, we get
$$
(a, b, c)=\left( \pm \frac{\sqrt{3}}{3}, \pm \frac{\sqrt{3}}{3}, \pm \frac{\sqrt{3}}{3}\right) .
$$

In summary, the required triples $(a, b, c)$ are $\left( \pm \frac{\sqrt{3}}{3}, \pm \frac{\sqrt{3}}{3}, \pm \frac{\sqrt{3}}{3}\right),(0, \pm 1, \pm 1)$ and their permutations.