How do you simplify #(4xy)^-2#?

1 Answer
EZ as pi
Jul 9, 2016

#1/(16x^2y^2)#

Explanation:

The negative index means that the whole bracket can be written in the denominator first to get rid of the minus. (Find the reciprocal)

#(4xy)^-2 = 1/(4xy)^2#

Now square everything in the bracket.

#1/(4xy)^2 = 1/(16x^2y^2)#

The alternative was to square each factor in the numerator, but the indices will all be negative, and have to be moved to the denominator separately.

The first method is probably quicker and easier.

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