Chapter 7 - Exponents and Exponential Functions - 7-5 Division Properties of Exponents - Practice and Problem-Solving Exercises - Page 444: 46

$\frac{1}{8n^{3}}$

Given : $(\frac{2n^{2-1}}{1}$$)^{-3}=$ $(\frac{2n}{1}$$)^{-3}=$ $2^{-3}\times n^{-3}=$ $\frac{1}{8}\times$ $\frac{1}{n^{3}}=$ $\frac{1}{8n^{3}}$ (since $(ab)^{m}=$ $a^{m}b^{m}$ and $a^{-m}=$ $\frac{1}{a^{m}}$)