How do you simplify #(25x^3)^(2/3) div (125xy^2) ^(1/3)#? Algebra Exponents and Exponential Functions Exponential Properties Involving Quotients 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)# Answer link Related questions What is the quotient of powers property? How do you simplify expressions using the quotient rule? What is the power of a quotient property? How do you evaluate the expression #(2^2/3^3)^3#? How do you simplify the expression #\frac{a^5b^4}{a^3b^2}#? How do you simplify #((a^3b^4)/(a^2b))^3# using the exponential properties? How do you simplify #\frac{(3ab)^2(4a^3b^4)^3}{(6a^2b)^4}#? Which exponential property do you use first to simplify #\frac{(2a^2bc^2)(6abc^3)}{4ab^2c}#? How do you simplify #(x^5y^8)/(x^4y^2)#? How do you simplify #[(2^3 *-3^2) / (2^4 * 3^-2)]^2#? See all questions in Exponential Properties Involving Quotients Impact of this question 1634 views around the world You can reuse this answer Creative Commons License