Determine the triplets of real numbers $(a, b, c)$ such that $a\left(b^{2}+c\right)=c(c+a b), b\left(c^{2}+a\right)=a(a+b c)$ and $c\left(a^{2}+b\right)=b(b+a c)$

## A square is positive
The equalities can be rewritten as $a b(b-c)=c(c-a)$, $b c(c-a)=a(a-b)$, and $c a(a-b)=b(b-c)$. By multiplying these three equalities, we obtain $(a b c)^{2}(c-a)(b-c)(a-b)=a b c(c-a)(b-c)(a-b)$. In particular, three cases arise:

- Either $a b c=0$. Since the equalities are cyclic, assume $a=0$. By the first equality, $c^{2}=0$ hence $c=0$. By the third equality, $b^{2}=0$ hence $b=0$ thus $a=b=c$.
- Or two of the elements among $a, b, c$ are equal, assume $a=b$. In this case, $b c(c-a)=0$. If $b=0$ or $c=0$, we fall back to the previous case, otherwise $a=b=c$.
- Or $a b c=1$, in which case $a, b, c$ are non-zero. The equalities can be rewritten as: $b-c=c^{2}(c-a)$, $c-a=a^{2}(a-b)$, and $a-b=b^{2}(b-c)$. In particular, $a-b$, $b-c$, and $c-a$ have the same sign, but their sum is zero, and we necessarily have $a-b=b-c=c-a=0$ i.e., $a=b=c$.

Conversely, if $a=b=c$, the different equations are indeed satisfied.