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

Sep 11, 2016

$2 {x}^{4} y$

#### Explanation:

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.

${\left(2 {x}^{-} 2 {y}^{\frac{1}{3}}\right)}^{\textcolor{red}{- 3}} {\left({16}^{\frac{1}{2}} {x}^{-} 1 y\right)}^{\textcolor{b l u e}{2}}$

=${2}^{-} 3 {x}^{6} {y}^{- \frac{3}{3}} \times {16}^{\frac{2}{2}} {x}^{-} 2 {y}^{2} \text{ "larr "sort out the negative indices}$

=$\frac{{x}^{6} \times 16 {y}^{2}}{{2}^{3} {x}^{2} y}$

=$\frac{{\cancel{16}}^{2} {x}^{6} {y}^{2}}{\cancel{8} {x}^{2} y} \text{ "larr " simplfy}$

=$2 {x}^{4} y$