How do you simplify #(2x^-2 y^(1/3))^-3(16^(1/2) x^-1 y)^2# using only positive exponents?

Question

1220 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-11T17:10:55

#2x^4y#

Wow there's a lot happening in here!

First, let's get rid of the brackets by multiplying the indices in each bracket by the powers to which they are raised.

#(2x^-2 y^(1/3))^color(red)(-3)(16^(1/2) x^-1 y)^color(blue)(2)#

=#2^-3x^6y^(-3/3) xx 16^(2/2)x^-2y^2" "larr "sort out the negative indices"#

=#(x^6 xx 16 y^2)/(2^3x^2y)#

=#(cancel16^2x^6y^2)/(cancel8x^2y) " "larr " simplfy"#

=#2x^4y#