Algebra 1

$\frac{1}{8n^{3}}$
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 $a^{-m}=$ $\frac{1}{a^{m}}$)