3)-$$\frac{2}{3}$$a²b⁶.

3) $$\frac{-2}{3}a^2b^6 = \frac{-2}{3}(ab^3)^2 = \frac{-2}{3}(\frac{4}{3})^2 = \frac{-2}{3} \cdot \frac{16}{9} = \frac{-32}{27}$$.

Ответ: $$\frac{-32}{27}$$