Respuesta :

(4x−2y^3xy−4)−2
=x6y^8/16y^6

=x^6y^2/16

Hope this helps:)
[tex]\bf ~~~~~~~~~~~~\textit{negative exponents}\\\\ a^{-n} \implies \cfrac{1}{a^n} \qquad \qquad \cfrac{1}{a^n}\implies a^{-n} \qquad \qquad a^n\implies \cfrac{1}{a^{-n}}\\\\ -------------------------------\\\\[/tex]

[tex]\bf (4x^{-2}y^3 \div xy^{-4})^{-2}\implies \left( \cfrac{4x^{-2}y^3}{xy^{-4}} \right)^{-2}\implies \left( \cfrac{xy^{-4}}{4x^{-2}y^3} \right)^{2} \\\\\\ \textit{then we distribute the exponent}\left( \cfrac{x^2y^{-4\cdot 2}}{4^2x^{-2\cdot 2}y^{3\cdot 2}} \right)\implies \cfrac{x^2y^{-8}}{16x^{-4}y^6} \\\\\\ \cfrac{x^2\cdot x^4}{16y^6\cdot y^8}\implies \cfrac{x^{2+4}}{16y^{6+8}}\implies \cfrac{x^6}{16y^{14}}[/tex]