Решение:

в)

$$\left(-\frac{2a^4}{3b^2}\right)^3 = - \frac{(2a^4)^3}{(3b^2)^3} = - \frac{2^3 (a^4)^3}{3^3 (b^2)^3} = - \frac{8a^{4\times3}}{27b^{2\times3}} = - \frac{8a^{12}}{27b^6}$$

г)

$$\frac{x^3 - 16x}{3xy} \cdot \frac{6y}{2x + 8} = \frac{x(x^2 - 16)}{3xy} \cdot \frac{6y}{2(x + 4)} = \frac{x(x - 4)(x + 4)}{3xy} \cdot \frac{6y}{2(x + 4)} = \frac{x(x - 4)(x + 4) \cdot 6y}{3xy \cdot 2(x + 4)} = \frac{6xy(x - 4)(x + 4)}{6xy(x + 4)} = x - 4$$

Ответ:

в) $$-\frac{8a^{12}}{27b^6}$$

г) $$x - 4$$