## Intermediate Algebra (6th Edition)

$\frac{27}{a^{12}b^{21}}$
We are given the expression $(\frac{3a^{-4}}{b^{7}})^{3}$. We can use the power of a quotient rule to simplify, which holds that $(\frac{a}{b})^{n}=\frac{a^{n}}{b^{n}}$, $b\ne0$ (where a and b are real numbers, and n is an integer). $(\frac{3a^{-4}}{b^{7}})^{3}=\frac{(3a^{-4})^{3}}{(b^{7})^{3}}$ To simplify further, we can use the power rule, which holds that $(a^{m})^{n}=a^{m\times n}$ (where a is a real number, and m and n are integers). $\frac{(3a^{-4})^{3}}{(b^{7})^{3}}=\frac{3^{3}a^{-4\times3}}{b^{7\times3}}=\frac{27a^{-12}}{b^{21}}=\frac{27}{a^{12}b^{21}}$