Solve the equation $\frac{(x^2+a^2-b^2)(a+x)}{ax} + \frac{(a^2+b^2-x^2)(b+a)}{ab} + \frac{(b^2+x^2-a^2)(x+b)}{bx} = 0$.

1. Problem Description

Solve the equation $\frac{(x^2+a^2-b^2)(a+x)}{ax} + \frac{(a^2+b^2-x^2)(b+a)}{ab} + \frac{(b^2+x^2-a^2)(x+b)}{bx} = 0$.

2. Solution Steps

First, multiply the entire equation by $abx$ to eliminate the denominators:

$b(x^2+a^2-b^2)(a+x) + x(a^2+b^2-x^2)(b+a) + a(b^2+x^2-a^2)(x+b) = 0$

Expanding each term:

$b(ax^2+x^3+a^3+a^2x-ab^2-b^2x) + x(a^2b+a^3+b^3+ab^2-bx^2-ax^2) + a(b^2x+bx^2+x^3+bx^2-a^2x-a^2b) = 0$

$abx^2+bx^3+a^3b+a^2bx-ab^3-b^3x + a^2bx+a^3x+b^3x+ab^2x-bx^3-ax^3 + ab^2x+abx^2+ax^3+abx^2-a^3x-a^3b = 0$

Combining like terms:

$(bx^3-bx^3) + (-ax^3+ax^3) + (abx^2+abx^2+abx^2) + (a^3b-a^3b) + (a^2bx+a^2bx+ab^2x+ab^2x) + (-ab^3) + (-b^3x+b^3x) + (a^3x-a^3x)=0$

$3abx^2 + 2a^2bx + 2ab^2x -ab^3 = 0$

$3abx^2 + 2ab(a+b)x - ab^3 = 0$

Divide by $ab$:

$3x^2 + 2(a+b)x - b^2 = 0$

Using the quadratic formula to solve for $x$:

$x = \frac{-2(a+b) \pm \sqrt{(2(a+b))^2 - 4(3)(-b^2)}}{2(3)}$

$x = \frac{-2(a+b) \pm \sqrt{4(a^2+2ab+b^2) + 12b^2}}{6}$

$x = \frac{-2(a+b) \pm \sqrt{4a^2+8ab+4b^2 + 12b^2}}{6}$

$x = \frac{-2(a+b) \pm \sqrt{4a^2+8ab+16b^2}}{6}$

$x = \frac{-2(a+b) \pm \sqrt{4(a^2+2ab+4b^2)}}{6}$

$x = \frac{-2(a+b) \pm 2\sqrt{a^2+2ab+4b^2}}{6}$

$x = \frac{-(a+b) \pm \sqrt{a^2+2ab+4b^2}}{3}$

3. Final Answer

$x = \frac{-(a+b) \pm \sqrt{a^2+2ab+4b^2}}{3}$