How do you simplify and write #((x^2)^-3(x^-2) )/ (x^2)^-4# with positive exponents?

1 Answer
Alan P.
Apr 5, 2016

#1#

Explanation:

Given: #(color(brown)((x^2)^(-3))color(red)((x^-2)))/(color(blue)((x^2)^(-4)))#

#color(brown)((x^2)^-3) = x^-6 = 1/x^6#

#color(red)((x^-2)) = 1/x^2#

#1/(color(blue)((x^2)^-4))=1/x^-8=x^8#

#(color(brown)((x^2)^(-3))color(red)((x^-2)))/(color(blue)((x^2)^(-4)))=color(brown)((x^2)^-3)xxcolor(red)((x^-2))xx1/(color(blue)((x^2)^-4))#

#color(white)("XXXXXXXX")=1/x^6xx1/x^2xxx^8/1#

#color(white)("XXXXXXXX")=x^8/x^8#

#color(white)("XXXXXXXX")=1#

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