Question 112532
So we want to simplify ((x^2)^-3)*(x^-2)/((x^2)^-4).
First, multiply exponents in each term to get (x^-6)*(x^-2)/(x^-8).
Now, when you multiply two terms with the same base, add the exponents, so get (x^-8)/(x^-8).  We could jump to the fact that anything divided by itself is 1, so this is 1.  You could also first put terms with negative exponents in the opposite position- numerator or denominator to get x^8/x^8 = 1.