# How do you simplify (-27)^(-2/3)?

Mar 12, 2017

$\frac{1}{9}$

#### Explanation:

Since ${a}^{\frac{m}{n}} = \sqrt[n]{{a}^{m}}$

and

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

then

${\left(- 27\right)}^{- \frac{2}{3}} = {\left(- \frac{1}{27}\right)}^{\frac{2}{3}} = \sqrt[3]{{\left(- \frac{1}{27}\right)}^{2}}$

Since $27 = {3}^{3}$, you get

$\sqrt[3]{{\left(- \frac{1}{3}\right)}^{6}} = {\left(- \frac{1}{3}\right)}^{2} = \frac{1}{9}$

Mar 12, 2017

$\frac{1}{9}$

#### Explanation:

Consider the example ${x}^{- \frac{2}{3}}$ this is the same as $\frac{1}{x} ^ \left(\frac{2}{3}\right)$

Using this as our guide:

Write ${\left(- 27\right)}^{- \frac{2}{3}}$ as $\frac{1}{- 27} ^ \left(\frac{2}{3}\right)$

This is the same as $\frac{1}{\sqrt[3]{{\left(- 27\right)}^{2}}}$

But I have just spotted something!
Lets take advantage of the fact that $\left(- 3\right) \times \left(- 3\right) \times \left(- 3\right) = - 27$

So changing the order we do things, write as:

$\frac{1}{\sqrt[3]{- 27}} ^ 2 = \frac{1}{{\left(- 3\right)}^{2}} = \frac{1}{9}$