How do you simplify #[(2^3 -3^2) / (2^4 3^-2)]^2#?

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Answer 1 · 2015-03-31T22:33:14

Remember the general transformations:
#a^p/q^q = a^(p-q)#
and
#a^(-p) = 1/(a^p)#

The given expression can be regrouped as
#[ 2^3/2^4 * (- 3^2/3^(-2))]^2#

#= [1/2 * (- 3^4)]^2#

#= 3^8/(2^2)#

#= (8281)/4#

How do you simplify #[(2^3 *-3^2) / (2^4 * 3^-2)]^2#?