Answer
$$\frac{1}{3x^{4}y^{4}} + \frac{8x^3}{y^{6}}$$
Work Step by Step
$$(\frac{3x^4}{y^{-4}})^{-1}+ (\frac{2x}{y^2})^{3}$$
Simplify the first term: $(\frac{3x^4}{y^{-4}})^{-1}$
Recall the power rule: $(a^{m})^{n}=a^{mn}$
Thus,
$$(\frac{3x^4}{y^{-4}})^{-1} = \frac{3^{-1}x^{4\cdot-1}}{y^{-4\cdot-1}} = \frac{3^{-1}x^{-4}}{y^{4}}$$
Recall the negative exponent rule: $a^{−n}=\frac{1}{a^{n}}$ and $\frac{1}{a^{-n}} = a^{n}$
Thus,
$$\frac{3^{-1}x^{-4}}{y^{4}} = \frac{1}{3x^{4}y^{4}}$$
Simplify the second term: $(\frac{2x}{y^2})^{3}$
Recall the power rule: $(a^{m})^{n}=a^{mn}$
Thus,
$$(\frac{2x}{y^2})^{3} = \frac{2^{3}x^3}{y^{2\cdot3}} = \frac{8x^3}{y^{6}}$$
Rewrite the equation:
$$\frac{1}{3x^{4}y^{4}} + \frac{8x^3}{y^{6}}$$
