# How do you simplify ((2x^-5y^2) / (4x^3y^-4)) ^-3?

Nov 24, 2015

$8 {x}^{24} {y}^{- 18}$

#### Explanation:

Let's remember the power rules first:

${a}^{m} \cdot {a}^{n} = {a}^{m + n}$

${a}^{- m} = \frac{1}{a} ^ m$

${\left({a}^{m}\right)}^{n} = a \cdot \left(m \cdot n\right)$

${\left(a \cdot b\right)}^{m} = {a}^{m} \cdot {b}^{m}$

Now, let's use these rules to simplify your expression:

${\left(\frac{2 {x}^{- 5} {y}^{2}}{4 {x}^{3} {y}^{- 4}}\right)}^{- 3} = {\left(\frac{2}{4} \cdot {x}^{- 5} / {x}^{3} \cdot {y}^{2} / {y}^{- 4}\right)}^{- 3}$

$\textcolor{w h i t e}{\times \times \times \times \times} = {\left(\frac{1}{2} \cdot {x}^{- 5} \cdot {x}^{- 3} \cdot {y}^{2} \cdot {y}^{4}\right)}^{- 3}$

$\textcolor{w h i t e}{\times \times \times \times \times} = {\left(\frac{1}{2} \cdot {x}^{- 8} \cdot {y}^{6}\right)}^{- 3}$

$\textcolor{w h i t e}{\times \times \times \times \times} = {\left(\frac{1}{2}\right)}^{- 3} \cdot {\left({x}^{- 8}\right)}^{- 3} \cdot {\left({y}^{6}\right)}^{- 3}$

$\textcolor{w h i t e}{\times \times \times \times \times} = {2}^{3} \cdot {x}^{24} \cdot {y}^{- 18}$

$\textcolor{w h i t e}{\times \times \times \times \times} = 8 {x}^{24} {y}^{- 18}$

$\textcolor{w h i t e}{\times \times \times \times \times} = \frac{8 {x}^{24}}{y} ^ 18$

Hope that this helped!