5. For the distinct real numbers $a, b, c$, it holds that

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

Calculate

$$
\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}
$$
Solution. We introduce the substitutions

$$
x=\frac{a}{b-c}, y=\frac{b}{c-a}, z=\frac{c}{a-b}
$$

Then the condition of the problem takes the form $x^{2}+y^{2}+z^{2}=2$, and the sum we are looking for is $x+y+z$. Let us note that

$$
\begin{gathered}
x+1=\frac{a+b-c}{b-c}, y+1=\frac{b+c-a}{c-a}, z+1=\frac{c+a-b}{a-b} \text { and } \\
x-1=\frac{a+c-b}{b-c}, y-1=\frac{b+a-c}{c-a}, z-1=\frac{c+b-a}{a-b}
\end{gathered}
$$

from which it follows that

$$
(x+1)(y+1)(z+1)=(x-1)(y-1)(z-1)
$$

The last equality is equivalent to the equality $x y+y z+z x=-1$. Then

$$
(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2(x y+y z+z x)=2-2=0
$$

from which we get $x+y+z=0$.