3. For non-zero real numbers $a, b, c$ it holds that

$$
a^{2}(b+c)=b^{2}(c+a)=c^{2}(a+b) .
$$

Determine all possible values of the expression

$$
\frac{(a+b+c)^{2}}{a^{2}+b^{2}+c^{2}}
$$

(Jaromír Šimša)
SOLUTION. We will show that the numbers 1 and 3 are the only possible values of the given expression. Note that due to the non-zero nature of the numbers $a, b, c$, we have $a^{2}+b^{2}+c^{2}>0$, so the given fraction makes sense and our calculations of its value will be correct.

First, let's manipulate the first of the given equations:

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

By similar manipulation of the second equation, we get

$$
(b-c)(a b+b c+c a)=0
$$

We see that if $a b+b c+c a \neq 0$, then $a-b=0$ and $b-c=0$, which means $a=b=c$. Therefore, it is necessary that $a b+b c+c a=0$ or $a=b=c$. Let's look at the value of the given expression in each of these two cases.

In the case $a b+b c+c a=0$, we have

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

while in the case $a=b=c$, we get

$$
\frac{(a+b+c)^{2}}{a^{2}+b^{2}+c^{2}}=\frac{(3 a)^{2}}{a^{2}+a^{2}+a^{2}}=\frac{9 a^{2}}{3 a^{2}}=3 .
$$

Both cases are possible, for example, the first occurs for $(a, b, c)=(3,6,-2)$ and the second for $(a, b, c)=(1,1,1)$ (Verification of the equations from the problem is easy for the triplet $(3,6,-2)$ and trivial for the triplet $(1,1,1)$.)[^1]