Respuesta :

apologiabiology
I can do both

remember
(ab)/(cd)=(a/c)(b/d)=(a/d)(b/c)

also exponential laws
(x^m)/(x^n)=x^(m-n)

and
(x^m)^n=x^(mn)
and
x^-m=1/(x^m)
and
(xyz)^m=(x^m)(y^m)(z^m)

first apply the exponents outside

(3x^-1y)^-2=(3^-2)((x^-1)^-2)(y^-2)=(1/9)x^2y^-2

bottom
(2x^2y^-2)^3=(2^3)((x^2)^3)((y^-2)^3)=8x^6y^-6

now we have
[tex] \frac{ (1/9)x^{2}y^{-2}}{8x^{6}y^{-6}} [/tex]
that equals
[tex]( \frac{1/9}{8} )( \frac{x^{2}}{x^{6}})( \frac{y^{-2}}{y^{-6}} ) [/tex] then
2-6=-4
-2-(-6)=4
[tex]( \frac{1}{72} )( \frac{1}{x^{4}})( \frac{1}{y^{4}} )[/tex]=
[tex] \frac{y^{4}}{72x^{4}} [/tex]

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