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 equality.

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 equality.

2. Solution Steps

First, let's simplify the second term denominator:

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

So, the given 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 for the left side:

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

So we want to prove:

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

This implies:

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

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

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

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

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

$3x = a+b$ (assuming $a+b \ne 0$).

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

However, the given problem asks to verify the equation. It looks like there is a typo in the original question.

Let us try to modify the right-hand side to make the problem true.

The equation

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

should be equal to $\frac{2}{(x+a+b)^2}$, which means we want to manipulate $\frac{2x}{(x-a-b)^2(x+a+b)}$ to equal $\frac{2}{(x+a+b)^2}$.

If the RHS were $\frac{2x}{(x+a+b)(x^2-(a+b)^2)} = \frac{2x}{(x+a+b)(x-a-b)(x+a+b)} = \frac{2x}{(x-a-b)(x+a+b)^2}$, the equation we want to prove is:

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

Multiplying both sides by $(x-a-b)^2(x+a+b)^2$ gives

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

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

We have $\frac{2x}{(x-a-b)^2(x+a+b)}$. We want this equal to something.

Let the RHS be $\frac{2}{x^2-a^2 - b^2 - 2ab} = \frac{2}{(x-a-b)(x+a+b)}$. Then:

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

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

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

$x+a+b = x-a-b$ which gives $2a+2b = 0$, which is false.

The original equation is unlikely to be correct.

3. Final Answer

The given equation is not generally true. It holds only when $x = (a+b)/3$ and $a+b \ne 0$. Therefore, there is no solution to the problem without modification or further context.

Final Answer: The equation is not generally true.