4. In the field of real numbers, solve the system of equations

$$
\begin{aligned}
& a\left(b^{2}+c\right)=c(c+a b) \\
& b\left(c^{2}+a\right)=a(a+b c), \\
& c\left(a^{2}+b\right)=b(b+c a) .
\end{aligned}
$$

(Michal Rolínek)
Solution. We will rearrange each equation so that the third-degree terms are on the left and the second-degree terms are on the right. After factoring out common factors, we obtain the equivalent system

$$
\begin{aligned}
& a b(b-c)=c(c-a) \\
& b c(c-a)=a(a-b) \\
& c a(a-b)=b(b-c)
\end{aligned}
$$

First, we investigate the possibility that two of the unknowns have the same value. If, for example, $a=b$, then from the third equation, we get either $b=c$ or $a=b=0$, and subsequently from the first equation, $c=0$. In both cases, $a=b=c$. The same result is obtained for any other pair of unknowns. Substitution easily verifies that the triplet $(t, t, t)$ is a solution to the given system for any real $t$.

If any of the unknowns is zero, for instance, $a=0$, it follows from the first equation that $c=0$ and subsequently from the third equation that $b=0$. Similarly, all three unknowns are zero if $b=0$ or if $c=0$. The triplet $(0,0,0)$ is already included among the solutions from the previous paragraph.

Finally, assume that the triplet $(a, b, c)$ is a solution, with no two of the numbers $a, b, c$ being the same and none of them being zero, i.e., all factors in all equations in (1) are non-zero. After multiplying these equations and dividing by common factors, we get $a b c=1$, from which, after multiplying each equation in (1) by the terms $c$, $a$, $b$ respectively, we obtain the equations

$$
\begin{aligned}
& b-c=c^{2}(c-a), \\
& c-a=a^{2}(a-b), \\
& a-b=b^{2}(b-c) .
\end{aligned}
$$

The factors $a^{2}, b^{2}, c^{2}$ are positive, so all three expressions $a-b, b-c, c-a$ have the same sign. However, this is not possible because their sum is zero. Therefore, no further solutions exist, and the only solutions to the given system are the triplets $(t, t, t), t \in \mathbb{R}$.