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 $x$

Question

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

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

Answer 1

I have a partial solution, as follows:

Note that $\frac{(b-c)(1+a^2)}{x+a^2}=\frac{(b-c)\left((x+a^2)+(1-x)\right)}{x+a^2}=(b-c)+\frac{{(b-c)}(1-x)}{x+a^2}$. Likewise, $\frac{(c-a)(1+b^2)}{x+b^2}=(c-a)+\frac{(c-a)(1-x)}{x+b^2}$ and $\frac{(a-b)(1+c^2)}{x+c^2}=(c-a)+\frac{(a-b)(1-x)}{x+c^2}$.

Now, $\frac{(b-c)(1+a^2)}{x+a^2}+\frac{(c-a)(1+b^2)}{x+b^2}+\frac{(a-b)(1+c^2)}{x+c^2}=\frac{(b-c)(1-x)}{x+a^2}+\frac{(c-a)(1-x)}{x+b^2}+\frac{(a-b)(1-x)}{x+c^2}$ as $(b-c)+(c-a)+(a-b)=0$.

Hence $(1-x)\left(\frac{b-c}{x+a^2}+\frac{c-a}{x+b^2}+\frac{a-b}{x+c^2}\right)=0$ and so $x=1$ or $\frac{b-c}{x+a^2}+\frac{c-a}{x+b^2}+\frac{a-b}{x+c^2}=0$

Answer 2

$$\frac{(b-c)(1+a^2)}{x+a^2}+\frac{(c-a)(1+b^2)}{x+b^2}+\frac{(a-b)(1+c^2)}{x+c^2}=0$$ $$\Leftrightarrow (b-c)(1+a^2)+\frac{(x+a^2)(c-a)(1+b^2)}{x+b^2}+\frac{(x+a^2)(a-b)(1+c^2)}{x+c^2}=0$$ $$\Leftrightarrow (b-c)(1+a^2)(x+b^2)+(x+a^2)(c-a)(1+b^2)+\frac{(x+a^2)(x+b^2)(a-b)(1+c^2)}{x+c^2}=0$$ $$\Leftrightarrow (b-c)(1+a^2)(x+b^2)(x+c^2)+(x+c^2)(x+a^2)(c-a)(1+b^2)+(x+a^2)(x+b^2)(a-b)(1+c^2)=0$$ $$\Leftrightarrow (b-c)(1+a^2)(x^2+x(b^2+c^2)+(bc)^4)+(x^2+x(a^2+c^2)+(ac)^4)(c-a)(1+b^2)+(x^2+x(b^2+a^2)+(ba)^4)(a-b)(1+c^2)=0$$ $$\Leftrightarrow (b-c)(1+a^2)(x^2)+x(b-c)(1+a^2)(b^2+c^2)+(bc)^4(b-c)(1+a^2)+(x^2)(c-a)(1+b^2)+x(c-a)(1+b^2)(a^2+c^2)+(ac)^4(c-a)(1+b^2)+(x^2)(a-b)(1+c^2)+x(b^2+a^2)(a-b)(1+c^2)+(ba)^4(a-b)(1+c^2)=0$$ $$\Leftrightarrow [(a-b)(1+c^2)+(b-c)(1+a^2)+(c-a)(1+b^2)](x^2)+x[(b^2+a^2)(a-b)(1+c^2)+(b-c)(1+a^2)(b^2+c^2)+(b^2+a^2)(a-b)(1+c^2)]+(bc)^4(b-c)(1+a^2)+(ac)^4(c-a)(1+b^2)(ba)^4(a-b)(1+c^2)=0$$ $$\Leftrightarrow x^2+\frac{(b^2+a^2)(a-b)(1+c^2)+(b-c)(1+a^2)(b^2+c^2)+(b^2+a^2)(a-b)(1+c^2)}{(a-b)(1+c^2)+(b-c)(1+a^2)+(c-a)(1+b^2)}x+\frac{(bc)^4(b-c)(1+a^2)+(ac)^4(c-a)(1+b^2)(ba)^4(a-b)(1+c^2)}{(a-b)(1+c^2)+(b-c)(1+a^2)+(c-a)(1+b^2)}=0$$

You can now simplify and solve the quadratic.

Answer 3

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)

Answer 4

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 $S = \left\{ {1;ab + bc + ca} \right\}$

Answer 5

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 $$x=1$$