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

Jan 11, 2017

$\frac{1}{16 {x}^{14} {y}^{9}}$

#### Explanation:

Before we start note that $2$ is the same as ${2}^{1}$

$\textcolor{b l u e}{\text{Consider } {x}^{- 4}}$

${x}^{- 4}$ is the same as $\frac{1}{x} ^ 4$

As an additional point: $\frac{1}{x} ^ \left(- 4\right)$ is the same as ${x}^{4}$
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color(blue)("Consider "(2x^3y^2)^3

This is the same as $\left(2 {x}^{3} {y}^{2}\right) \times \left(2 {x}^{3} {y}^{2}\right) \times \left(2 {x}^{3} {y}^{2}\right)$

Giving:$\text{ } 8 {x}^{9} {y}^{6}$

So as a general case:$\text{ } {\left({x}^{a} {y}^{b}\right)}^{c} = {x}^{a \times c} {y}^{b \times c}$
So for ${\left(2 {x}^{3} {y}^{2}\right)}^{3} = {2}^{1 \times 3} {x}^{3 \times 3} {y}^{2 \times 3}$

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$\textcolor{b l u e}{\text{Putting it all together}}$
${x}^{-} \frac{4}{2 x {y}^{3} {\left(2 {x}^{3} {y}^{2}\right)}^{3}} = \frac{1}{x} ^ 4 \times \frac{1}{2 x {y}^{3} {\left(2 {x}^{3} {y}^{2}\right)}^{3}} = \frac{1}{x} ^ 4 \times \frac{1}{2 x {y}^{3} \times 8 {x}^{9} {y}^{6}}$

$\text{ } = \frac{1}{16 {x}^{14} {y}^{9}}$