Question

1. $\frac{a^{2}-b^{2}}{a^{3}-b^{3}}cdot\frac{a^{2}+ab + b^{2}}{a^{2}+2ab + b^{2}}$

Explanation:

Step1: Factor the expressions

We know that \(a^{2}-b^{2}=(a + b)(a - b)\), \(a^{3}-b^{3}=(a - b)(a^{2}+ab + b^{2})\) and \(a^{2}+2ab + b^{2}=(a + b)^{2}\).
So the original - expression \(\frac{a^{2}-b^{2}}{a^{3}-b^{3}}\cdot\frac{a^{2}+ab + b^{2}}{a^{2}+2ab + b^{2}}\) becomes \(\frac{(a + b)(a - b)}{(a - b)(a^{2}+ab + b^{2})}\cdot\frac{a^{2}+ab + b^{2}}{(a + b)^{2}}\).

Step2: Cancel out the common factors

Cancel out the common factors \((a - b)\) and \((a^{2}+ab + b^{2})\) and one factor of \((a + b)\) in the numerator and denominator.
\(\frac{(a + b)(a - b)}{(a - b)(a^{2}+ab + b^{2})}\cdot\frac{a^{2}+ab + b^{2}}{(a + b)^{2}}=\frac{1}{a + b}\)

Answer:

\(\frac{1}{a + b}\)