Let’s solve the given expression step by step:
Given:
[tex]\[
\frac{(a+b)^2}{a-b} \cdot \frac{a^3-b^3}{a^2-b^2} \div \frac{a^2+a b+b^2}{(a-b)^2}
\][/tex]
Step 1: Simplify each expression separately.
1. Simplify \(\frac{(a+b)^2}{a-b}\).
[tex]\[
\frac{(a+b)^2}{a-b} = \frac{a^2 + 2ab + b^2}{a-b}
\][/tex]
2. Simplify \(\frac{a^3 - b^3}{a^2 - b^2}\).
Recall the factorization formulas:
[tex]\[
a^3 - b^3 = (a - b)(a^2 + ab + b^2)
\][/tex]
[tex]\[
a^2 - b^2 = (a - b)(a + b)
\][/tex]
Therefore:
[tex]\[
\frac{a^3 - b^3}{a^2 - b^2} = \frac{(a - b)(a^2 + ab + b^2)}{(a - b)(a + b)} = \frac{a^2 + ab + b^2}{a + b}
\][/tex]
3. Simplify \(\frac{a^2 + ab + b^2}{(a - b)^2}\).
[tex]\[
\frac{a^2 + ab + b^2}{(a - b)^2}
\][/tex]
Step 2: Substitute these simplified expressions back into the original problem.
The original problem:
[tex]\[
\frac{(a+b)^2}{a-b} \cdot \frac{a^3 - b^3}{a^2 - b^2} \div \frac{a^2 + ab + b^2}{(a - b)^2}
\][/tex]
becomes:
[tex]\[
\frac{a^2 + 2ab + b^2}{a-b} \cdot \frac{a^2 + ab + b^2}{a + b} \div \frac{a^2 + ab + b^2}{(a - b)^2}
\][/tex]
Step 3: Convert the division to multiplication by the reciprocal:
[tex]\[
\frac{a^2 + 2ab + b^2}{a-b} \cdot \frac{a^2 + ab + b^2}{a + b} \cdot \frac{(a - b)^2}{a^2 + ab + b^2}
\][/tex]
Step 4: Cancel out the common terms.
The terms \(a^2 + ab + b^2\) in the numerator and denominator will cancel out:
[tex]\[
\frac{a^2 + 2ab + b^2}{a-b} \cdot \frac{(a - b)^2}{a + b}
\][/tex]
Now, it simplifies to:
[tex]\[
\frac{a^2 + 2ab + b^2}{a-b} \cdot \frac{(a - b)}{(a + b)}
\][/tex]
[tex]\[
\frac{a^2 + 2ab + b^2}{(a + b)}
\][/tex]
Finally:
[tex]\[
a^2 + 2ab + b^2
\][/tex]
The result is:
[tex]\[
a^2 - b^2
\][/tex]
Thus, the final simplified result is:
[tex]\[
\boxed{a^2 - b^2}
\][/tex]
