# How do you simplify (root3(-162x^6))?

Oct 3, 2015

$\sqrt[3]{- 162 {x}^{6}} = {x}^{2} \left(- 3 \sqrt[3]{6}\right)$

#### Explanation:

We can take a minus out of the root, because $- 1 = {\left(- 1\right)}^{3}$

$- \sqrt[3]{162 {x}^{6}}$

We know that ${x}^{6} = {x}^{2} \cdot {x}^{2} \cdot {x}^{2}$ or ${x}^{6} = {\left({x}^{2}\right)}^{3}$, so we can take a ${x}^{2}$ out of the root

$- {x}^{2} \sqrt[3]{162}$

Lastly, all we have to do is factor the $162$, so

$162 | 2$
$\textcolor{w h i t e}{0} 81 | 3$
$\textcolor{w h i t e}{0} 27 | 3$
$\textcolor{w h i t e}{00} 9 | 3$
$\textcolor{w h i t e}{00} 3 | 3$
$\textcolor{w h i t e}{00} 1 | 2 \cdot {3}^{4}$

So we can take a $3$ out of the root, leaving a $2 \cdot 3$, that is, $6$ inside it. So

$\sqrt[3]{- 162 {x}^{6}} = {x}^{2} \left(- 3 \sqrt[3]{6}\right)$