Answer
$$8x^{3}y^{9}$$
Work Step by Step
$$\frac{7x^3}{y^{-9}}+ (\frac{x^{-1}}{y^3})^{-3}$$
Simplify the first term: $\frac{7x^3}{y^{-9}}$
Recall the negative exponent rule: $a^{−n}=\frac{1}{a^{n}}$ and $\frac{1}{a^{-n}} = a^{n}$
Thus,
$$\frac{7x^3}{y^{-9}} = 7x^{3}y^{9}$$
Simplify the second term: $(\frac{x^{-1}}{y^3})^{-3}$
Recall the power rule: $(a^{m})^{n}=a^{mn}$
Thus,
$$(\frac{x^{-1}}{y^3})^{-3} = \frac{x^{-1\cdot-3}}{y^{3\cdot-3}} = \frac{x^{3}}{y^{-9}}$$
Recall the negative exponent rule: $a^{−n}=\frac{1}{a^{n}}$ and $\frac{1}{a^{-n}} = a^{n}$
Thus,
$$\frac{x^{3}}{y^{-9}} = x^{3}y^{9}$$
Rewrite the equation:
$$7x^{3}y^{9} + x^{3}y^{9} = 8x^{3}y^{9}$$
