# Simplify (125/729)^(-1/3)?

Oct 3, 2016

${\left(\frac{125}{729}\right)}^{- \frac{1}{3}} = \frac{9}{5}$

#### Explanation:

We should use the identities ${\left({a}^{m}\right)}^{n} = {a}^{m \times n}$, ${a}^{- m} = \frac{1}{a} ^ m$ and ${a}^{\frac{1}{m}} = \sqrt[m]{a}$

Hence, ${\left(\frac{125}{729}\right)}^{- \frac{1}{3}}$

= ${\left(\frac{5 \times 5 \times 5}{3 \times 3 \times 3 \times 3 \times 3 \times 3}\right)}^{- \frac{1}{3}}$

= ${\left({5}^{3} / {3}^{6}\right)}^{- \frac{1}{3}}$

= 5^((3xx(-1/3)))/3^((6xx(-1/3))

= ${5}^{- 1} / {3}^{- 2}$

= $\frac{\frac{1}{5}}{\frac{1}{9}}$

= $\frac{1}{5} \times \frac{9}{1}$

= $\frac{9}{5}$

Oct 3, 2016

$\frac{9}{5}$

#### Explanation:

Recall: ${x}^{-} m = \frac{1}{x} ^ m \text{ and } {\left(\frac{\textcolor{red}{x}}{\textcolor{b l u e}{y}}\right)}^{-} m = {\left(\frac{\textcolor{b l u e}{y}}{\textcolor{red}{x}}\right)}^{m}$

I usually try to get rid of any negative indices first.
Use the second law shown above to do this.

${\left(\frac{125}{729}\right)}^{- \frac{1}{3}} \text{ = } {\left(\frac{729}{125}\right)}^{\frac{1}{3}}$

It is useful to learn all the powers up to 1000.

You should recognise these values as being cubes.
(${5}^{3} = 125 \mathmr{and} {9}^{3} = 729$)

Recall: $\sqrt[3]{x} \Leftrightarrow {x}^{\frac{1}{3}}$

${\left(\frac{729}{125}\right)}^{\frac{1}{3}} = \sqrt[3]{\left(\frac{729}{125}\right)} = \frac{\sqrt[3]{729}}{\sqrt[3]{125}} = \frac{\sqrt[3]{{9}^{3}}}{\sqrt[3]{{5}^{3}}}$

$= \frac{9}{5}$