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 $x \in \{1;ab+bc+ca\}$

Multiplying the original equation out is
$-a^3b^2x + a^3c^2x + a^2b^3x - a^2c^3x - b^3c^2x + b^2c^3x + a^3b^2 - a^3c^2 - a^2b^3 +a^2bx^2 + a^2c^3 - a^2cx^2 - ab^2x^2 + ac^2x^2 + b^3c^2 - b^2c^3 + b^2cx^2 - bc^2x^2 a^2bx + a^2cx + ab^2x - ac^2x - b^2cx + bc^2x=0$
Taking all the coefficients $a$, $b$ and $c$ in the quadratic $ax^2+bx+c=0$ grouping them  and then factoring leads to the following quadratic $$(b-c)(a-c)(a-b)x^2-(b-c)(a-c)(a-b) (ab+ac+bc+1)x+(b-c)(a-c)(a-b)(ab+ac+bc)=0$$ Dividing both sides of the equation by $$(b-c)(a-c)(a-b)$$ gives $$x^2-(ab+ac+bc+1)x+(ab+ac+bc)=0$$ Using the quadratic formula
$$x=\frac{(ab+ac+bc+1)\pm{\sqrt{(ab + ac + bc + 1)^2-4(ab + ac + bc)}}}{2}$$ or $$x=\frac{(ab+ac+bc+1)\pm\sqrt{(ab + ac + bc-1)^2}}{2}$$ So the roots are $$x=ab+ac+bc$$ and $$x=1$$
