Simplify #((x^-1y^4)/(x^2y^7))^-3# ?

1 Answer
Feb 21, 2017

Answer:

#((x^-1y^4)/(x^2y^7))^-3=x^9y^9#

Explanation:

The expression as written in the (original - before restoration) question is ambiguous as it is unclear whether denominator should be read as the product of the #x^2# and #y^7# terms or only the #x^2#.

I have chosen to interpret the expression as #((x^-1y^4)/(x^2y^7))^-3#

Two rules of indices will apply here:

(i) #a^m xx a^n = a^(m+n)#
(ii) #(a^m)^n = a^(m xx n)#

#((x^-1y^4)/(x^2y^7))^-3 = (x^(-1-2) xx y^(4-7))^-3# [Rule (i)]

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

#=x^9y^9# [Rule (ii)]