# How do you simplify (64n^12)^(-1/6)?

Jan 16, 2017

See the entire simplification process below:

#### Explanation:

First, we will use these rule of exponents to start the simplification process:

${\left({x}^{\textcolor{red}{a}}\right)}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} \times \textcolor{b l u e}{b}}$

$a = {a}^{\textcolor{red}{1}}$

${\left(64 {n}^{\textcolor{red}{12}}\right)}^{\textcolor{b l u e}{- \frac{1}{6}}} = {\left({64}^{\textcolor{red}{1}} {n}^{\textcolor{red}{12}}\right)}^{\textcolor{b l u e}{- \frac{1}{6}}} =$

${64}^{\textcolor{red}{1} \times \textcolor{b l u e}{- \frac{1}{6}}} {n}^{\textcolor{red}{12} \times \textcolor{b l u e}{- \frac{1}{6}}} = {64}^{- \frac{1}{6}} {n}^{-} 2$

Now we can use this rule for exponents to continue the simplification process:

${x}^{\textcolor{red}{a}} = \frac{1}{x} ^ \textcolor{red}{- a}$

${64}^{- \frac{1}{6}} {n}^{-} 2 = \frac{1}{{64}^{- - \frac{1}{6}} {n}^{- - 2}} = \frac{1}{{64}^{\frac{1}{6}} {n}^{2}}$

Now, to finish we need to know ${2}^{6} = 64$ and ${\left(- 2\right)}^{6} = 64$ therefore ${64}^{\frac{1}{6}} = 2$ or $- 2$

$\frac{1}{{64}^{\frac{1}{6}} {n}^{2}} = \frac{1}{2 {n}^{2}}$ or $\frac{1}{{64}^{\frac{1}{6}} {n}^{2}} = - \frac{1}{2 {n}^{2}}$