Problem 1. Determine all triples $(a, b, c)$ of real numbers for which the following relations hold simultaneously:

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

Czech and Slovak Olympiad
Solution: Let $(a, b, c)$ be a solution to the system. If one of the numbers $a, b, c$ is equal to 0, for example, if $c=0$, then from $c\left(a^{2}+b\right)=a(b+c a)$ it follows that $a=0$, and similarly, that $b=0$. Therefore, if $a b c=0$, then $a=b=c=0$. We continue to look for solutions with $a b c \neq 0$. The relations from the statement can be rewritten equivalently as

\[
\begin{aligned}
& a b(b-c)=c(c-a) \\
& b c(c-a)=a(a-b) \\
& c a(a-b)=b(b-c) .
\end{aligned}
\]

Thus, $a^{2} b^{2} c^{2}(a-b)(b-c)(c-a)=a b c(a-b)(b-c)(c-a)$.

Case 1: Among the numbers $a, b, c$ there are (at least) two equal; let these be $a$ and $b$. We have

\[
a^{2} c+b c=b^{2}+a b c \Leftrightarrow b c=b^{2} \Leftrightarrow c=b
\]

so in this case we get $a=b=c$. Conversely, any triplet with equal components satisfies the given relations.

Case 2: If $a \neq b \neq c \neq a$, then $a b c=1$. It follows that $a b(b-c)=c(c-a) \Leftrightarrow(b-c)=c^{2}(c-a)$ and two similar relations. It follows that

\[
a^{3}+b^{3}+c^{3}=a c^{2}+b a^{2}+c b^{2}
\]

If $a, b, c>0$, then from the inequality of means we have $a^{3}+a^{3}+b^{3} \geq 3 b a^{2}$ (with equality if $a=b$) which, added to its analogues, leads to $a^{3}+b^{3}+c^{3} \geq a c^{2}+b a^{2}+c b^{2}$. We thus have equality in the previous inequality, so $a=b=c$.

If one of the variables is positive and the other two are negative, for example $a>0, b, c<0$, then

\[
a^{3}+b^{3}+c^{3} \geq a c^{2}+b a^{2}+c b^{2} \Rightarrow a^{3}+b^{3}+c^{3} \geq 0 \Rightarrow a<0
\]

a contradiction.

In conclusion, the only solutions are $(a, b, c)=(x, x, x), \forall x \in \mathbb{R}$.