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[(2^3 *-3^2) / (2^4 * 3^-2)]^2?