We are given the equation: $\frac{1}{(x-a-b)(x+a+b)} + \frac{1}{x^2 + a^2 + b^2 - 2ax - 2bx + 2ab} = \frac{2}{(x+a+b)^2}$. We need to verify this equation.

1. Problem Description

We are given the equation:

$\frac{1}{(x-a-b)(x+a+b)} + \frac{1}{x^2 + a^2 + b^2 - 2ax - 2bx + 2ab} = \frac{2}{(x+a+b)^2}$.

We need to verify this equation.

2. Solution Steps

First, let's simplify the denominator of the second term on the left side.

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

So the equation becomes:

$\frac{1}{(x-a-b)(x+a+b)} + \frac{1}{(x-a-b)^2} = \frac{2}{(x+a+b)^2}$.

Now, let's find a common denominator on the left side, which is $(x-a-b)^2(x+a+b)$.

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

So, we have:

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

Cross-multiplying, we get:

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

Dividing by 2,

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

Assuming $x+a+b \neq 0$, we can divide both sides by $x+a+b$.

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

Expanding both sides:

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

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

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

$3ax + 3bx = a^2 + 2ab + b^2$.

$3x(a+b) = (a+b)^2$.

Assuming $a+b \neq 0$, we can divide both sides by $a+b$.

$3x = a+b$.

$x = \frac{a+b}{3}$.

However, the equation $\frac{2x}{(x-a-b)^2(x+a+b)} = \frac{2}{(x+a+b)^2}$ does not generally hold true. For the equation to be true for all $x, a, b$, it means that there is an error in the original problem or the equation given.

The provided equation is:

$\frac{1}{(x-a-b)(x+a+b)} + \frac{1}{x^2 + a^2 + b^2 - 2ax - 2bx + 2ab} = \frac{2}{(x+a+b)^2}$

$\frac{1}{(x-a-b)(x+a+b)} + \frac{1}{(x-a-b)^2} = \frac{2}{(x+a+b)^2}$.

This is not an identity.

3. Final Answer

The given equation is not an identity. There appears to be an error in the problem.