Exercise 5. Find all triplets of real numbers $(\mathrm{a}, \mathrm{b}, \mathrm{c})$ satisfying the system of equalities:

$$
\left\{\begin{array}{l}
a\left(b^{2}+c\right)=c(c+a b) \\
b\left(c^{2}+a\right)=a(a+b c) \\
c\left(a^{2}+b\right)=b(b+a c)
\end{array}\right.
$$
Solution to Exercise 5 The equalities can be rewritten as $ab(b-c)=c(c-a)$, $bc(c-a)=a(a-b)$, and $ca(a-b)=b(b-c)$. By multiplying these three equalities, we obtain $(abc)^2(c-a)(b-c)(a-b) = abc(c-a)(b-c)(a-b)$. In particular, two cases arise:

- Either $\mathrm{abc}=0$. Since the equalities are cyclic, assume $\mathrm{a}=0$. By the first equality, $\mathrm{c}^2=0$ so $\mathrm{c}=0$. By the third equality, $\mathrm{b}^2=0$ so $\mathrm{b}=0$.
- Or two of the elements among $a, b, c$ are equal, assume $a=b$. In this case, $bc(c-a)=0$. If $\mathrm{b}=0$ or $\mathrm{c}=0$, we fall back to the previous case; otherwise, $\mathrm{a}=\mathrm{b}=\mathrm{c}$.
- Or $abc=1$, 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, so we must have $a-b=b-c=c-a=0$, i.e., $a=b=c$.

Comment from the graders: Many students submitted a solution to this problem, but few received full marks. Indeed, there were several recurring errors:

- If $\mathrm{a} \geqslant \mathrm{b}$ and $\mathrm{c} \geqslant \mathrm{d}$, it is not necessarily true that $\mathrm{ac} \geqslant \mathrm{bd}$. This is true for positive real numbers, but not for negative real numbers: for example, $2 \geqslant -1, 0 \geqslant -1$ but $0 \times 2=0 < (-1) \times (-1)=1$.
- Several students assumed $\mathrm{a} \geqslant \mathrm{b} \geqslant \mathrm{c}$. However, such an assumption cannot be made since the system is not symmetric: swapping $a$ and $b$ completely changes the system. On the other hand, the system is cyclic: replacing $\mathrm{a}$ with $\mathrm{b}$, $\mathrm{b}$ with $\mathrm{c}$, and $\mathrm{c}$ with $\mathrm{a}$ does not change the system. Therefore, we can assume that $a$ is the largest, but we cannot assume any additional inequality on $b$ and $c$.
- Some students forgot to verify that $\mathrm{a}=\mathrm{b}=\mathrm{c}$ is a solution: it is necessary to do so.