What is #(x^-6y^9)^(1/3)#?

1 Answer
May 2, 2016

#y^3/x^2#

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#