How do you simplify #(25x^3)^(2/3) div (125xy^2) ^(1/3)#?

1 Answer
EZ as pi
Sep 6, 2016

#x^2root3((5)/(xy^2)#

Explanation:

Recall: laws of indices:
#x^(1/3) = root3 x#
#x^m xx y^m = (xy)^m#

#(25x^3)^(2/1 xx1/3) div (125xy^2) ^(1/3)#

= #((25x^3)^(2/1 xxcolor(red)(1/3)))/ ((125xy^2) ^color(red)(1/3)) = " "[((25x^3)^(2/1))/(125xy^2)] ^color(red)(1/3)#

#[(25xx25x^6)/(125xy^2)] ^color(red)(1/3) = " " [(cancel25xxcancel25^5x^6)/(cancel125^cancel5xy^2)] ^color(red)(1/3)#

=#root3((5x^6)/(xy^2)#

=#x^2root3((5)/(xy^2)#

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