$$
\begin{array}{l}
\text { Find all ordered triples of real numbers } (a, b, c): \\
\quad a^{2} b+c=b^{2} c+a=c^{2} a+b, \\
a b+b c+c a=1 .
\end{array}
$$
1. (1) If one of $a, b, c$ is 0, let's assume $a=0$, then
$$
b c=1, c=b^{2} c=b \Rightarrow b=c= \pm 1 .
$$

Similarly, $b=0, a=c= \pm 1$;
$$
c=0, a=b= \pm 1 \text {. }
$$

Upon verification, all of the above satisfy the conditions.
(2) If $a, b, c$ are all non-zero, from
$$
\begin{array}{l}
b^{2} c-c=a^{2} b-a=a(a b-1) \\
=a(-b c-c a)=-a b c-c a^{2} \\
\Rightarrow a b+a^{2}+b^{2}=1 .
\end{array}
$$

Similarly, $b c+b^{2}+c^{2}=1$,
$$
c a+c^{2}+a^{2}=1 \text {. }
$$

Adding the above three equations, we get
$$
a b+b c+c a+2\left(a^{2}+b^{2}+c^{2}\right)=3 \text {. }
$$

Given $a b+b c+c a=1$, we know $a^{2}+b^{2}+c^{2}=1$.
$$
\begin{array}{l}
\text { Also, }(a-b)^{2}+(b-c)^{2}+(c-a)^{2} \\
=2\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)=0,
\end{array}
$$

then $a=b=c= \pm \frac{\sqrt{3}}{3}$.
$$
\begin{array}{l}
\text { Combining }(1) \text { and }(2), (a, b, c) \text { are } \\
(0,1,1),(0,-1,-1),(1,1,0), \\
(-1,-1,0),(1,0,1),(-1,0,-1) \text {, } \\
\left(\frac{\sqrt{3}}{3}, \frac{\sqrt{3}}{3}, \frac{\sqrt{3}}{3}\right),\left(-\frac{\sqrt{3}}{3},-\frac{\sqrt{3}}{3},-\frac{\sqrt{3}}{3}\right) .
\end{array}
$$