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

My solution:
Now $\left( {b - c} \right)\left( {1 + {a^2}} \right) = {a^2}\left( {b - c} \right) + \left( {b - c} \right) + \left( {b - c} \right)x - \left( {b - c} \right)x = \left( {b - c} \right)\left( {x + {a^2}} \right) + \left( {b - c} \right)\left( {1 - x} \right)$.
Then $\dfrac{{\left( {b - c} \right)\left( {1 + {a^2}} \right)}}{{x + {a^2}}} = \dfrac{{\left( {b - c} \right)\left( {x + {a^2}} \right) + \left( {b - c} \right)\left( {1 - x} \right)}}{{x + {a^2}}} = b - c + \dfrac{{\left( {1 - x} \right)\left( {b - c} \right)}}{{x + {a^2}}}$.
Same, $\dfrac{{\left( {c - a} \right)\left( {1 + {b^2}} \right)}}{{x + {b^2}}} = c - a + \dfrac{{\left( {1 - x} \right)\left( {c - a} \right)}}{{x + {b^2}}}\,\,\,\,\& \,\,\,\dfrac{{\left( {a - b} \right)\left( {1 + {c^2}} \right)}}{{x + {c^2}}} = a - b + \dfrac{{\left( {1 - x} \right)\left( {a - b} \right)}}{{x + {c^2}}}$
Because $\left( {a - b} \right) + \left( {b - c} \right) + \left( {c - a} \right) = 0$ so the equation becomes 
$\left( {1 - x} \right)\left( {\dfrac{{b - c}}{{x + {a^2}}} + \dfrac{{c - a}}{{x + {b^2}}} + \dfrac{{a - b}}{{x + {c^2}}}} \right) = 0 \Longleftrightarrow \left[ \begin{array}{l}
x = 1\\
\dfrac{{b - c}}{{x + {a^2}}} + \dfrac{{c - a}}{{x + {b^2}}} + \dfrac{{a - b}}{{x + {c^2}}} = 0\left( \bigstar \right)
\end{array} \right.$
$\left( \bigstar \right) \Longleftrightarrow  - \dfrac{{\left( {c - a} \right) + \left( {a - b} \right)}}{{x + {a^2}}} + \dfrac{{c - a}}{{x + {b^2}}} + \dfrac{{a - b}}{{x + {c^2}}} = 0 \Longleftrightarrow \dfrac{{c - a}}{{x + {b^2}}} - \dfrac{{c - a}}{{x + {a^2}}} + \dfrac{{a - b}}{{x + {c^2}}} - \dfrac{{a - b}}{{x + {a^2}}} = 0$
$\Longleftrightarrow \left( {c - a} \right)\left( {\dfrac{1}{{x + {b^2}}} - \dfrac{1}{{x + {a^2}}}} \right) + \left( {a - b} \right)\left( {\dfrac{1}{{x + {c^2}}} - \dfrac{1}{{x + {a^2}}}} \right) = 0$
$\Longleftrightarrow \dfrac{{\left( {c - a} \right)\left( {a - b} \right)\left( {a + b} \right)}}{{\left( {x + {a^2}} \right)\left( {x + {b^2}} \right)}} - \dfrac{{\left( {a - b} \right)\left( {c - a} \right)\left( {a + c} \right)}}{{\left( {x + {a^2}} \right)\left( {x + {c^2}} \right)}} = 0 \Longleftrightarrow \dfrac{{\left( {a - b} \right)\left( {c - a} \right)}}{{x + {a^2}}}\left( {\dfrac{{a + b}}{{x + {b^2}}} - \dfrac{{a + c}}{{x + {c^2}}}} \right) = 0$
$\Longleftrightarrow \dfrac{{a + b}}{{x + {b^2}}} = \dfrac{{a + c}}{{x + {c^2}}} \Longleftrightarrow \left( {a + b} \right)x + {c^2}\left( {a + b} \right) = x\left( {a + c} \right) + {b^2}\left( {a + c} \right)$
$\Longleftrightarrow x\left( {b - c} \right) = a\left( {{b^2} - {c^2}} \right) + bc\left( {b - c} \right) \Longleftrightarrow x = a\left( {b + c} \right) + bc \Longleftrightarrow x = ab + bc + ca$.
Hence $S = \left\{ {1;ab + bc + ca} \right\}$
