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, we rewrite the second term denominator:

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

Then we have the left-hand side (LHS) as

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

Find a common denominator which is $(x-a-b)^2(x+a+b)$. Then,

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

The equation we are trying to prove can be rewritten as:

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

Let us examine this equation. We simplify the left-hand side by adding the two fractions.

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

The problem suggests that this is equal to $\frac{2}{(x+a+b)^2}$. This is not correct.

However, if the problem statement has a typo, and we assume that the second fraction has a numerator of 1, then,

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

Combining the two terms on the left, we get:

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

This does not equal $\frac{2}{(x+a+b)^2}$.

There might be a typo in the problem statement.

Let's consider an example:

$a=1, b=1, x=5$. Then the original equation is:

$\frac{1}{(5-1-1)(5+1+1)} + \frac{1}{5^2+1^2+1^2-2(1)(5)-2(1)(5)+2(1)(1)} = \frac{2}{(5+1+1)^2}$

$\frac{1}{(3)(7)} + \frac{1}{25+1+1-10-10+2} = \frac{2}{49}$

$\frac{1}{21} + \frac{1}{9} = \frac{2}{49}$

$\frac{3+7}{63} = \frac{10}{63} \ne \frac{2}{49}$

The given equation is incorrect.

3. Final Answer

The given equation is incorrect. There must be a typo in the problem statement.