How do you simplify #(3x^3y)/((3y)^-2)#?

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Answer 1 · 2018-05-10T19:58:22

#27x^3y^3#

#1/(3y)^(-2)=(3y)^2#

#(3x^3y)/(3y)^(-2)=3x^3y xx (3y)^2#

#=3x^3y xx 9y^2=27x^3y^3#

Answer 2 · 2018-05-10T20:02:10

#(3x^3y)/(3y)^(-2)# = #=27x^3y^3# = #(3xy)^3#

You want to simplify #(3x^3y)/(3y)^(-2)#

We must remember that #1/x^-n=x^n#
Therefore #1/(3y)^(-2)=(3y)^2#

We, therefore, can write:
#(3x^3y)/(3y)^(-2)#
#=(3x^3y)(3y)^2#
#=3x^3y*3^2y^2#
#=27x^3y^3#

As #27=3^3# it's perhaps a little more elegant if we write this
#(3xy)^3#