Problem: Find all triples $(a, b, c)$ of real numbers such that $ab + bc + ca = 1$ and
\[ a^2b + c = b^2c + a = c^2a + b. \]

Solution: Answer: $(\pm 1/\sqrt3, \pm 1/\sqrt3, \pm1/\sqrt3)$
where the signs correspond,
and $(\pm1, \pm1, 0)$ and permutations where the signs correspond.
These work and we prove that is all.

We begin by eliminating the condition
via homogenization:
the first equality now reads
\begin{align*}
  a^2b+c\left[ ab+bc+ca \right] &= b^2c+a\left[ ab+bc+ca \right] \\
  \iff c^2(b+a) &= c(b^2+a^2) \\
  \iff c&=0 \text{ or } a^2+b^2 = c(a+b).
\end{align*}
Cyclic variations hold.
So we have two cases.

\begin{itemize}
\ii If any of the variables is zero,
say $a = 0$, then the other two are nonzero.
So from $b^2 = bc$ we get $b = c$ giving
$(\pm1, \pm1, 0)$.

\ii Now assume all three variables are nonzero,
so $a^2+b^2 = c(a+b)$.
If we sum cyclically we get
\[ 2(a^2+b^2+c^2) = 2(ab+bc+ca)
  \iff (a-b)^2 + (b-c)^2 + (c-a)^2 = 0 \]
which forces $a=b=c$ and gives the last solution.
\end{itemize}