What is #(x^-6y^9)^(1/3)#?
1 Answer
May 2, 2016
Explanation:
I am assuming you mean in simplified form with positive indices.
Using the following
#color(blue)" rules of exponents "#
#• (a^m)^n=a^(mn)" and " a^-m hArr 1/a^m # example:
#(2^3)^2=2^(3xx2)=2^6=64# and
#2^-3=1/2^3=1/8 #
#rArr(x^-6y^9)^(1/3)=x^(-6xx1/3)y^(9xx1/3)=x^-2y^3=y^3/x^2#