(Netherlands). Find all triples $(a, b, c)$ of real numbers such that $a b+b c+$ $c a=1$ and

$$
a^{2} b+c=b^{2} c+a=c^{2} a+b
$$
First suppose that \(a=0\). Then we have \(bc=1\) and \(c=b^2c=b\). So \(b=c\), which implies \(b^2=1\) and hence \(b=\pm 1\). This leads to the solutions \((a, b, c)=(0,1,1)\) and \((a, b, c)=(0,-1,-1)\). Similarly, \(b=0\) gives the solutions \((a, b, c)=(1,0,1)\) and \((a, b, c)=(-1,0,-1)\), while \(c=0\) gives \((a, b, c)=(1,1,0)\) and \((a, b, c)=(-1,-1,0)\).

Now we may assume that \(a, b, c \neq 0\). We multiply \(ab + bc + ca = 1\) by \(a\) to find \(a^2b + abc + ca^2 = a\), hence \(a^2b = a - abc - a^2c\). Substituting this in \(a^2b + c = b^2c + a\) yields \(a - abc - a^2c + c = b^2c + a\), so \(b^2c + abc + a^2c = c\). As \(c \neq 0\), we find \(b^2 + ab + a^2 = 1\).

Analogously we have \(b^2 + bc + c^2 = 1\) and \(a^2 + ac + c^2 = 1\). Adding these three equations yields \(2(a^2 + b^2 + c^2) + ab + bc + ca = 3\), which implies \(a^2 + b^2 + c^2 = 1\). Combining this result with \(b^2 + ab + a^2 = 1\), we get \(1 - ab = 1 - c^2\), so \(c^2 = ab\).

Analogously we also have \(b^2 = ac\) and \(a^2 = bc\). In particular we now have that \(ab, bc\), and \(ca\) are all positive. This means that \(a, b\), and \(c\) must all be positive or all be negative. Now assume that \(|c|\) is the largest among \(|a|, |b|\), and \(|c|\), then \(c^2 \geq |ab| = ab = c^2\), so we must have equality. This means that \(|c| = |a|\) and \(|c| = |b|\). Since \((a, b, c)\) must all have the same sign, we find \(a = b = c\). Now we have \(3a^2 = 1\), hence \(a = \pm \frac{1}{3} \sqrt{3}\). We find the solutions \((a, b, c) = \left(\frac{1}{3} \sqrt{3}, \frac{1}{3} \sqrt{3}, \frac{1}{3} \sqrt{3}\right)\) and \((a, b, c) = \left(-\frac{1}{3} \sqrt{3}, -\frac{1}{3} \sqrt{3}, -\frac{1}{3} \sqrt{3}\right)\).

We conclude that all possible triples \((a, b, c)\) are \((0,1,1), (0,-1,-1), (1,0,1), (-1,0,-1)\), \((1,1,0), (-1,-1,0), \left(\frac{1}{3} \sqrt{3}, \frac{1}{3} \sqrt{3}, \frac{1}{3} \sqrt{3}\right)\) and \(\left(-\frac{1}{3} \sqrt{3}, -\frac{1}{3} \sqrt{3}, -\frac{1}{3} \sqrt{3}\right)\).