Question

I really don’t understand this and just need an explanation of how to do it.

Answer 1

$\bf ~\hspace{12em}\left( \cfrac{2n}{-3n\cdot -2n^2} \right)^4 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \cfrac{2n}{-3n\cdot -2n^2}\implies \cfrac{1}{-3n}\cdot \cfrac{2n}{-2n^2}\implies \cfrac{1}{-3n}\cdot \cfrac{2n}{2n\cdot -n}\implies \cfrac{1}{-3n}\cdot \cfrac{2n}{2n}\cdot \cfrac{1}{-n}$

$\bf \cfrac{1}{-3n}\cdot \boxed{1}\cdot \cfrac{1}{-n}\implies \cfrac{1}{-3n\cdot -n}\implies \cfrac{1}{3n^2} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \left( \cfrac{2n}{-3n\cdot -2n^2} \right)^4\implies \left( \cfrac{1}{3n^2} \right)^4\implies \stackrel{\textit{distributing the exponent}}{\cfrac{1^4}{3^4n^{2\cdot 4}}}\implies \cfrac{1}{81n^8}$