4) 3b²+6b+3/b³-8 * 2b²+4b+8/9b+9.

4) $$\frac{3b^2+6b+3}{b^3-8} \cdot \frac{2b^2+4b+8}{9b+9} = \frac{3(b^2+2b+1)}{(b-2)(b^2+2b+4)} \cdot \frac{2(b^2+2b+4)}{9(b+1)} = \frac{3(b+1)^2}{(b-2)(b^2+2b+4)} \cdot \frac{2(b^2+2b+4)}{9(b+1)} = \frac{6(b+1)^2(b^2+2b+4)}{9(b-2)(b+1)(b^2+2b+4)} = \frac{2(b+1)}{3(b-2)}$$

Ответ: 2(b+1)/(3(b-2))