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

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

What values can the expression

$$
p=\frac{(a+b)(b+c)(c+a)}{a b c}
$$

take?
(1) implies that none of $a, b$, and $c$ is zero, so (2) is well-defined. Let's add 2 to each term of (1)! We get that (1) is true if and only if

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

This holds if and only if either $a+b+c=0$ or $a=b=c$. In the first case, $a+b=-c$, $b+c=-a$, and $c+a=-b$. Substituting these into (2) gives $p=-1$. In the second case, $p=8$. Therefore, the value of ( $p$ ) is either -1 or 8, and both values are clearly attainable.