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Simplify: [tex]\frac{2b}{(a+b)^2(a-b)} - \frac{1}{a^2 - b^2} + \frac{1}{a^2 + 2ab + b^2}[/tex]

Sagot :

To simplify the given expression, we start by analyzing each term individually and looking for common patterns or identities that we can utilize. The expression is:

[tex]\[ \frac{2 b}{(a+b)^2(a-b)} - \frac{1}{a^2 - b^2} + \frac{1}{a^2 + 2ab + b^2} \][/tex]

First, let's rewrite each term by recognizing possible factorizations:

#### Term 1:
[tex]\[ \frac{2 b}{(a+b)^2(a-b)} \][/tex]

#### Term 2:
[tex]\[ \frac{1}{a^2 - b^2} \][/tex]

Notice that [tex]\(a^2 - b^2\)[/tex] can be factored using the difference of squares:
[tex]\[ a^2 - b^2 = (a + b)(a - b) \][/tex]

So the second term becomes:
[tex]\[ \frac{1}{a^2 - b^2} = \frac{1}{(a+b)(a-b)} \][/tex]

#### Term 3:
[tex]\[ \frac{1}{a^2 + 2ab + b^2} \][/tex]

Notice that [tex]\(a^2 + 2ab + b^2\)[/tex] is a perfect square:
[tex]\[ a^2 + 2ab + b^2 = (a + b)^2 \][/tex]

So the third term becomes:
[tex]\[ \frac{1}{a^2 + 2ab + b^2} = \frac{1}{(a+b)^2} \][/tex]

Now we substitute the factored forms back into the original expression:

[tex]\[ \frac{2 b}{(a+b)^2(a-b)} - \frac{1}{(a+b)(a-b)} + \frac{1}{(a+b)^2} \][/tex]

Next, let's find a common denominator for these fractions. The common denominator will be [tex]\((a+b)^2(a-b)\)[/tex].

Rewriting each fraction with the common denominator:

[tex]\[ \frac{2 b}{(a+b)^2(a-b)} - \frac{(a+b)}{(a+b)^2(a-b)} + \frac{(a-b)}{(a+b)^2(a-b)} \][/tex]

Now the expression is:

[tex]\[ \frac{2 b - (a+b) + (a-b)}{(a+b)^2(a-b)} \][/tex]

Simplify the numerator:

[tex]\[ 2 b - (a + b) + (a - b) = 2b - a - b + a - b = 0 \][/tex]

So, the expression simplifies to:

[tex]\[ \frac{0}{(a+b)^2(a-b)} = 0 \][/tex]

Thus, the simplified form of the given expression is:

[tex]\[ \boxed{0} \][/tex]