How do you simplify (8/27)^(-2/3)?

Feb 14, 2017

$\frac{9}{4}$

Explanation:

${\left(\frac{8}{27}\right)}^{- \frac{2}{3}}$
$\text{ }$
$= {\left(\frac{27}{8}\right)}^{\frac{2}{3}}$
$\text{ }$
The prime factorization of 27 and 8 is:
$\text{ }$
$27 = 9 \times 3 = 3 \times 3 \times 3 = {3}^{3}$
$\text{ }$
$8 = 4 \times 2 = 2 \times 2 \times 2 = {2}^{3}$

Substituting the factorization on the above fraction we have:
$\text{ }$
${\left(\frac{27}{8}\right)}^{\frac{2}{3}}$
$\text{ }$
$= {\left({3}^{3} / {2}^{3}\right)}^{\frac{2}{3}}$
$\text{ }$
$= {\left({\left(\frac{3}{2}\right)}^{3}\right)}^{\frac{2}{3}}$
$\text{ }$
Applying the property of power of a power that says:
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$\textcolor{red}{{\left({a}^{n}\right)}^{m} = {a}^{m \times n}}$
$\text{ }$
${\left({\left(\frac{3}{2}\right)}^{3}\right)}^{\frac{2}{3}}$
$\text{ }$
$= {\left(\frac{3}{2}\right)}^{3 \times \left(\frac{2}{3}\right)}$
$\text{ }$
$= {\left(\frac{3}{2}\right)}^{2}$
$\text{ }$
$= \frac{9}{4}$

Feb 14, 2017

$= \frac{9}{4}$

Explanation:

There are 3 different processes indicated in this expression.

Laws of indices:

${x}^{-} m = \frac{1}{x} ^ m \text{ "and" } {\left(\frac{a}{b}\right)}^{-} m = {\left(\frac{b}{a}\right)}^{m}$

The second law is the one we will apply.

Also ${x}^{\frac{p}{q}} = {\sqrt[q]{x}}^{p}$

${\left(\frac{8}{27}\right)}^{- \frac{2}{3}} = {\left(\frac{27}{8}\right)}^{+ \frac{2}{3}}$

$= {\sqrt[3]{\frac{27}{8}}}^{2} \text{ } \leftarrow$ find the cube roots first

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

$= \frac{9}{4}$