Solve $\frac{(b-c)(1+a^2)}{x+a^2}+\frac{(c-a)(1+b^2)}{x+b^2}+\frac{(a-b)(1+c^2)}{x+c^2}=0$ for $x$ Is there any smart way to solve the equation: $$\frac{(b-c)(1+a^2)}{x+a^2}+\frac{(c-a)(1+b^2)}{x+b^2}+\frac{(a-b)(1+c^2)}{x+c^2}=0$$

Use Maple I can find $x \in \{1;ab+bc+ca\}$

Starting with pipi's simplification of the mess:
$$
\frac{b - c}{x + a^2} + \frac{c - a}{x + b^2} + \frac{a - b}{x + c^2} = 0 \\
(b - c) (x + b^2) (x + c^2) + (c - a) (x + a^2) (x + c^2) + (a - b) (x + a^2) (x + b^2) = 0
$$
Suprisingly, this turns out a linear equation for $x$, with solution:
$$
x = a b + a c + b c
$$
(Many thanks to Maxima for help with algebra)
