2.1. Given real numbers \(a, b, c\) such that \(a+b+c=0\). Find

\[
\left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right)\left(\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c}\right)
\]
### 2.1. Answer: 9.

Notice that for three numbers $x, y, z$ the identity holds:

$$
\frac{x-z}{y}+\frac{z-y}{x}+\frac{y-x}{z}=\frac{(x-y)(y-z)(z-x)}{x y z}
$$

Apply this to the set $(x, y, z)=(b-c, c-a, a-b)$. Notice that $x-z=2 b-a-c=3 b$ and so on, so we get

$$
\frac{3 a}{b-c}+\frac{3 b}{c-a}+\frac{3 c}{a-b}=\frac{-27 a b c}{(b-c)(c-a)(a-b)}
$$

From this, we find that the first factor is $-(9 a b c)((b-c)(c-a)(a-b))$, and the second, by the same identity for the set $(x, y, z)=(a, b, c)$, is $-(b-c)(c-a)(a-b) a b c$. It remains to multiply and simplify.