To solve this problem efficiently, let's examine and simplify each fraction before performing the multiplication.

The expression given is:

$\frac{6a - 6b}{5a - 5b} \cdot \frac{2a + 2b}{a^2 - b^2}$

Step 1: Factorize Each Term

1. For the first fraction, $\frac{6a - 6b}{5a - 5b}$:

   • Factor out $6$ in the numerator: $6a - 6b = 6(a - b)$.
   • Factor out $5$ in the denominator: $5a - 5b = 5(a - b)$.
   • This gives us: \frac{6(a - b)}{5(a - b)} = \frac{6}{5} \quad \text{(since a - b cancels out)}

2. For the second fraction, $\frac{2a + 2b}{a^2 - b^2}$:

   • Factor out $2$ in the numerator: $2a + 2b = 2(a + b)$.
   • Recognize that $a^2 - b^2$ is a difference of squares: $a^2 - b^2 = (a + b)(a - b)$.
   • This gives us: \frac{2(a + b)}{(a + b)(a - b)} = \frac{2}{a - b} \quad \text{(since a + b cancels out)}

Step 2: Multiply the Simplified Fractions

Now, we have: $\frac{6}{5} \cdot \frac{2}{a - b} = \frac{12}{5(a - b)}$

Final Answer:

$\frac{12}{5(a - b)}$

Would you like a further breakdown of any step, or do you have additional questions?

Related Questions

1. What is the difference between simplifying a fraction and factoring it?
2. How do you handle more complex factorizations in algebraic expressions?
3. What other techniques can simplify expressions like this?
4. How do you know when terms can cancel in a fraction?
5. How does the difference of squares factorization work in general?

Tip: Always check for common factors first to simplify expressions quickly.