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

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3  
Clear the fractions, and it's a quadratic in $x$. –  Gerry Myerson Feb 17 '13 at 12:34
    
Good question, +1. The coefficients of the corresponding quadratic equation are atrocious, so I wouldn't go this way. Where did you find this question? –  1015 Feb 17 '13 at 13:34

4 Answers 4

up vote 10 down vote accepted

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 $\frac{b-c}{x+a^2}+\frac{c-a}{x+b^2}+\frac{a-b}{x+c^2}=0$

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Nice observation, +1. –  1015 Feb 17 '13 at 13:51

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

You can now simplify and solve the quadratic.

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5  
And I understand why you let the OP simplify and solve the quadratic... –  1015 Feb 17 '13 at 13:25
    
I really appreciate the effort you took to write down all these. –  dineshdileep Feb 17 '13 at 15:52
    
Dear George; If I've ever given a pity upvote, it's now! (Just teasing of course :)) –  Bruno Joyal Feb 23 '13 at 2:07
    
Well thanks, @Bruno. –  Student Feb 23 '13 at 7:19
    
No offense George, but the previous questions seem to answer the question more properly. I wouldn't call brute force a "smart way" to solve this equation even though it's a very legitimate way to do things. –  Patrick Da Silva Feb 24 '13 at 3:47

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)

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Surprisingly is the right word, +1. –  1015 Feb 17 '13 at 15:44
    
Well, linearity is not so suprising, the $x^2$ term is seen to have coefficient $(b - c) + (c - a) + (a - b) = 0$. And the overall symmetry makes a symmetrical solution probable (that's why I got interested ;-). –  vonbrand Feb 17 '13 at 15:51

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 $$x=1$$

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