For real numbers $a, b, c$ it holds that

$$
\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}
$$

Determine all possible values of the expression

$$
\frac{a^{3}+b^{3}+c^{3}}{(b+c)^{3}+(c+a)^{3}+(a+b)^{3}}.
$$

(Michal Rolínek)
SOLUTION. From the problem statement, it follows that the numbers $a, b, c$ satisfy the conditions

$$
b+c \neq 0, \quad c+a \neq 0, \quad a+b \neq 0 .
$$

Under the assumptions $(\mathrm{P})$, we equivalently transform the first of the given equalities:

$$
\begin{gathered}
\frac{a}{b+c}=\frac{b}{c+a}, \\
a(c+a)=b(b+c), \\
a c+a^{2}=b^{2}+b c \\
(a-b)(a+b)+c(a-b)=0, \\
(a-b)(a+b+c)=0 .
\end{gathered}
$$

By similar transformations of the second equality and also the equality of the third fraction with the first fraction, we get

$$
(b-c)(a+b+c)=0 \quad \text { and } \quad(c-a)(a+b+c)=0
$$

We see that under the assumptions $(\mathrm{P})$, the real numbers $a, b, c$ satisfy the problem statement if and only if $a+b+c=0$ or $a=b=c$.

In the case where $a+b+c=0$, for the given expression we have

$$
\frac{a^{3}+b^{3}+c^{3}}{(b+c)^{3}+(c+a)^{3}+(a+b)^{3}}=\frac{a^{3}+b^{3}+c^{3}}{(-a)^{3}+(-b)^{3}+(-c)^{3}}=-1,
$$

provided that $a^{3}+b^{3}+c^{3} \neq 0$, otherwise the expression makes no sense.

In the second case, where $a=b=c$, we have

$$
\frac{a^{3}+b^{3}+c^{3}}{(b+c)^{3}+(c+a)^{3}+(a+b)^{3}}=\frac{a^{3}+a^{3}+a^{3}}{(2 a)^{3}+(2 a)^{3}+(2 a)^{3}}=\frac{1}{8},
$$

but this time under the condition that $a^{3} \neq 0$.

It remains to show that each of the two found values is achievable.

The conditions $(\mathrm{P})$ and the equality $a+b+c=0$ are satisfied, for example, by the triplet $(a, b, c)=$ $=(1,1,-2)$, for which the expression $a^{3}+b^{3}+c^{3}$ (indeed non-zero) has the value 6.

The conditions $(\mathrm{P})$ and the equality $a=b=c$ are satisfied, for example, by the triplet $(a, b, c)=$ $=(1,1,1)$, for which the expression $a^{3}$ (indeed non-zero) has the value 1.

Conclusion. The given expression has only two possible values: -1 and $1 / 8$.

Comment. Without verifying the condition $a^{3}+b^{3}+c^{3} \neq 0$ for the chosen triplet $(1,1,-2)$, the provided solution would not be complete. Although more laborious, but ultimately more interesting, is to show that under the conditions $(\mathrm{P})$ never both equalities $a+b+c=0$ and $a^{3}+b^{3}+c^{3}=0$ hold. If the first of them holds, we can substitute $c=-a-b$ into the second equality and further transform its left side:

$a^{3}+b^{3}+c^{3}=a^{3}+b^{3}+(-a-b)^{3}=-3 a^{2} b-3 a b^{2}=-3 a b(a+b)=-3(b+c)(c+a)(a+b)$,

where in the last step we replaced the factors $a, b$ with $-b-c$, resp. $-c-a$. It remains to add that due to the conditions $(\mathrm{P})$, the product $(b+c)(c+a)(a+b)$ is not equal to zero.