We are asked to verify the given 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

We are asked to verify the given 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:

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

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

Expanding the numerator and denominator:

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

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

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

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

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

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

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

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

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

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

Now, let us express the numerator as follows:

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

However, we cannot obtain the numerator as $2(x^2+(a+b)^2)$.

Consider the case when $x=a+b$, then $LHS = \frac{2(2(a+b)^2)}{0}$ which is undefined.

Also, $RHS = \frac{2((a+b)^2 - a(a+b) - b(a+b) + ab) + 3(a^2+b^2)}{0} = \frac{2(a^2+b^2+2ab - a^2 - ab -ab -b^2 + ab)+3(a^2+b^2)}{0} = \frac{2ab+3(a^2+b^2)}{0}$. This expression is also undefined.

However, we cannot conclude whether the equation is true or false.

Expanding the RHS numerator, we have:

$2x^2 - 2ax - 2bx + 2ab + 3a^2 + 3b^2 = 2x^2 + 2(a^2 + 2ab + b^2) = 2x^2 + 2a^2 + 2b^2 + 4ab$ only if $-2ax -2bx+2ab+3a^2 + 3b^2 = 2a^2 + 2b^2 + 4ab$

This simplifies to $a^2+b^2-2ax-2bx +2ab = 0$, i.e., $(a+b)^2 = 2x(a+b)$, or $a+b = 2x$. In this case, the given equation holds. However, this must hold for all $x, a, b$. Therefore, this is not an identity.

The given identity is false in general.

3. Final Answer

The given equation is false.