# How do you simplify ((2x^-6)/(7y^-3))^-3?

Jan 9, 2016

$\frac{343 {x}^{18}}{8 {y}^{9}}$

#### Explanation:

A negative power ${x}^{- n}$ can be seen as ${x}^{{n}^{- 1}}$. We know that ${x}^{- 1} = \frac{1}{x}$ so ${x}^{{n}^{- 1}} = \frac{1}{x} ^ n$, hence the following simplyfictations : $2 {x}^{-} 6 = \frac{2}{x} ^ 6$ and $7 {y}^{-} 3 = \frac{7}{y} ^ 3$

So ${\left(\frac{2 {x}^{-} 6}{7 {y}^{-} 3}\right)}^{- 3} = {\left(\frac{\frac{2}{x} ^ 6}{\frac{7}{y} ^ 3}\right)}^{- 3} = {\left(\left(\frac{2}{x} ^ 6\right) \cdot \left({y}^{3} / 7\right)\right)}^{- 3}$.

We do what we did before : ${\left(\left(\frac{2}{x} ^ 6\right) \cdot \left({y}^{3} / 7\right)\right)}^{- 3} = {\left({x}^{6} / 2 \cdot \frac{7}{y} ^ 3\right)}^{3} = \frac{343 {x}^{18}}{8 {y}^{9}}$

Jan 9, 2016

A slightly different way or writing the working out!

$\frac{343 {x}^{18}}{8 {y}^{9}}$

#### Explanation:

Given: ${\left(\frac{2 {x}^{- 6}}{7 {y}^{- 3}}\right)}^{- 3}$

First consider inside the brackets:

Known that $\textcolor{w h i t e}{. .} 2 {x}^{- 6} \to \frac{2}{{x}^{6}} \text{ and that } \frac{1}{7 {y}^{- 3}} \to {y}^{3} / 7$

So we have $\frac{2}{x} ^ 6 \times {y}^{3} / 7 = \frac{2 {y}^{3}}{7 {x}^{6}}$

Now put this back into the brackets giving

${\left(\frac{2 {y}^{3}}{7 {x}^{6}}\right)}^{- 3} = {\left(\frac{7 {x}^{6}}{2 {y}^{3}}\right)}^{3}$

$= \frac{343 {x}^{18}}{8 {y}^{9}}$