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

3) $$\frac{5a+5b}{b} \cdot \frac{6b^2}{a^2-b^2} = \frac{5(a+b)}{b} \cdot \frac{6b^2}{(a-b)(a+b)} = \frac{5}{1} \cdot \frac{6b}{(a-b)} = \frac{30b}{a-b}$$

Ответ: $$\frac{30b}{a-b}$$