$$
\frac{a+b}{(b-c)(c-a)}+\frac{b+c}{(c-a)(a-b)}+\frac{c+a}{(a-b)(b-c)}=?
$$
When we bring the fractions to a common denominator, we then get the following numerator:

$$
(a+b)(a-b)+(b+c)(b-c)+(c+a)(c-a)=a^{2}-b^{2}+b^{2}-c^{2}+c^{2}-a^{2}=0
$$

thus, the value of the questioned sum is 0.

Number of solutions: 26.