# How do you simplify and write (4x^-4y^2)^-3 with positive exponents?

Jul 10, 2018

${x}^{12} / \left(64 {y}^{6}\right)$

#### Explanation:

Our expression is the same as

$\frac{1}{4 {x}^{-} 4 {y}^{2}} ^ 3$

${4}^{3} = \textcolor{b l u e}{64}$, ${\left({x}^{-} 4\right)}^{3} = {x}^{- 4 \cdot 3} = \textcolor{b l u e}{{x}^{- 12}}$, ${\left({y}^{2}\right)}^{3} - {y}^{2 \cdot 3} = \textcolor{b l u e}{{y}^{6}}$

With this in mind, we now have the following:

$\frac{1}{64 {x}^{-} 12 {y}^{6}}$

We can make the negative exponent positive by bringing it to the numerator. We now have

${x}^{12} / \left(64 {y}^{6}\right)$

Hope this helps!