remember some basic rules
\((ab)^c=(a^c)(b^c)\) and
\((a^b)^c=a^{bc}\) and
\(x^{-m}= \frac{1}{x^m}\)
so
\((3x^{-2}y^5)^4=((3x^{-2})^4)((y^5)^4)\)
\((3x^{-2})^4=(3^4)((x^{-2}^4)=(81)(x^{-8})\)
\((y^5)^4=y^{20}\)
we now have
\(81x^{-8}y^{20}\)
\(x^{-8}= \frac{1}{x^8}\) so
\(81x^{-8}y^{20}= \frac{81y^{20}}{x^8}\)
