The problem is to verify the following equation: $\frac{x+a+b}{x-a-b} + \frac{x-a-b}{x+a+b} = \frac{2(x^2 - ax - bx + ab) + 3(a^2 + b^2)}{x^2 - a^2 - 2ab - b^2}$

1. Problem Description

The problem is to verify the following equation:

$\frac{x+a+b}{x-a-b} + \frac{x-a-b}{x+a+b} = \frac{2(x^2 - ax - bx + ab) + 3(a^2 + b^2)}{x^2 - a^2 - 2ab - b^2}$

2. Solution Steps

Let's simplify the left-hand side (LHS) of the equation:

$LHS = \frac{x+a+b}{x-a-b} + \frac{x-a-b}{x+a+b}$

$LHS = \frac{(x+a+b)^2 + (x-a-b)^2}{(x-a-b)(x+a+b)}$

$LHS = \frac{x^2 + a^2 + b^2 + 2ax + 2bx + 2ab + x^2 + a^2 + b^2 - 2ax - 2bx + 2ab}{x^2 - (a+b)^2}$

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

$LHS = \frac{2x^2 + 2a^2 + 2b^2 + 4ab}{x^2 - a^2 - 2ab - b^2}$

$LHS = \frac{2(x^2 + a^2 + b^2 + 2ab)}{x^2 - a^2 - 2ab - b^2}$

$LHS = \frac{2(x^2 + (a+b)^2)}{x^2 - (a+b)^2}$

Now let's simplify the right-hand side (RHS) of the equation:

$RHS = \frac{2(x^2 - ax - bx + ab) + 3(a^2 + b^2)}{x^2 - a^2 - 2ab - b^2}$

$RHS = \frac{2x^2 - 2ax - 2bx + 2ab + 3a^2 + 3b^2}{x^2 - (a+b)^2}$

We want to show that $LHS = RHS$, or

$\frac{2(x^2 + a^2 + b^2 + 2ab)}{x^2 - a^2 - 2ab - b^2} = \frac{2x^2 - 2ax - 2bx + 2ab + 3a^2 + 3b^2}{x^2 - a^2 - 2ab - b^2}$

This is equivalent to:

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

$2x^2 + 2a^2 + 2b^2 + 4ab = 2x^2 - 2ax - 2bx + 2ab + 3a^2 + 3b^2$

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

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

However, we want to find an expression that is always true for all $x$, $a$, and $b$. This equation $a^2 + b^2 - 2ax - 2bx - 2ab = 0$ is not generally true. There must be some restrictions in the question for x, a, or b that are not stated.

The numerators can be simplified further as follows:

$2(x^2 + (a+b)^2) = 2x^2 + 2a^2 + 4ab + 2b^2$

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

Then

$2x^2 + 2a^2 + 4ab + 2b^2 = 2x^2 - 2ax - 2bx + 2ab + 3a^2 + 3b^2$

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

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

3. Final Answer

The equation is only true when $a^2 + b^2 - 2ax - 2bx - 2ab = 0$.