# How do you find the value of 125^(-2/3)?

Sep 23, 2015

${125}^{- \frac{2}{3}} = \frac{1}{25}$

#### Explanation:

${125}^{- \frac{2}{3}}$

$= \frac{1}{{125}^{\frac{2}{3}}}$

$= \frac{1}{{\left({125}^{\frac{1}{3}}\right)}^{2}}$

$= \frac{1}{{\left(5\right)}^{2}}$

$= \frac{1}{25}$

Sep 23, 2015

$\frac{1}{25}$

#### Explanation:

According to the law of indices, ${a}^{-} m = \frac{1}{a} ^ m$. So you only need to get the reciprocal to get rid of the negative exponent.

${125}^{- \frac{2}{3}}$
=1/(125^(2/3)

According to the law of indices again, ${a}^{\frac{m}{n}} = \sqrt[n]{{a}^{m}}$.

1/(125^(2/3)
$= \frac{1}{\sqrt[3]{{125}^{2}}}$

Let's factor out 125 completely so we can take out any perfect cubes.

$\frac{1}{\sqrt[3]{{125}^{2}}}$
$= \frac{1}{\sqrt[3]{{\left({5}^{3}\right)}^{2}}}$
$= \frac{1}{\sqrt[3]{{5}^{3} \cdot {5}^{3}}}$
$= \frac{1}{5 \cdot 5}$
$= \textcolor{b l u e}{\frac{1}{25}}$

Sep 26, 2016

$\frac{1}{25}$

#### Explanation:

Note that $125 = {5}^{3}$.

Thus:

${\left(125\right)}^{- \frac{2}{3}} = {\left({5}^{3}\right)}^{- \frac{2}{3}}$

Now we should use the rule ${\left({a}^{b}\right)}^{c} = {a}^{b \times c}$, so:

${\left({5}^{3}\right)}^{- \frac{2}{3}} = {5}^{\left(3 \times - \frac{2}{3}\right)} = {5}^{-} 2$

Here, use the rule ${a}^{-} b = \frac{1}{a} ^ b$:

${5}^{-} 2 = \frac{1}{5} ^ 2 = \frac{1}{25}$