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

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60
Jul 24, 2016

$= 4$

#### Explanation:

${8}^{\frac{2}{3}}$ can be written as
root3(8^2
$= \sqrt[3]{64}$
=root3((4)(4)(4)
$= 4$

Then teach the underlying concepts
Don't copy without citing sources
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#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

8
Jun 30, 2017

$4$

#### Explanation:

One of the laws of indices states: ${x}^{\frac{m}{n}} = \sqrt[n]{\left({x}^{m}\right)} = {\left(\sqrt[n]{x}\right)}^{m}$

Therefore ${8}^{\frac{2}{3}}$ can be written as $\sqrt[3]{\left({8}^{2}\right)} \mathmr{and} {\left(\sqrt[3]{8}\right)}^{2}$

I prefer to use the second form because it uses smaller numbers - the root is found first and then that is squared.

$\sqrt[3]{8} = 2 \mathmr{and} {2}^{2} = 4$

So: ${\left(\textcolor{b l u e}{\sqrt[3]{8}}\right)}^{2} = {\textcolor{b l u e}{2}}^{2} = 4$

Consider a question such as ${32}^{\frac{3}{5}}$

$\sqrt[5]{\textcolor{b l u e}{\left({32}^{3}\right)}}$ would mean finding ${32}^{3}$ first ... ouch!

${\left(\textcolor{b l u e}{\sqrt[5]{32}}\right)}^{3}$ would mean finding $\sqrt[5]{32}$ first. That is $\textcolor{b l u e}{2}$

${\left(\textcolor{b l u e}{\sqrt[5]{32}}\right)}^{3} = {\textcolor{b l u e}{2}}^{3} = 8$

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