# How do you simplify -root8(4)+5root8(4)?

Jun 12, 2017

See a solution process below:

#### Explanation:

First, rewrite this expression as:

$- 1 \sqrt[8]{4} + 5 \sqrt[8]{4}$

Next, factor and combine these like terms:

$\left(- 1 + 5\right) \sqrt[8]{4} = 4 \sqrt[8]{4}$

We can now rewrite this expression using this rule for exponents and radicals:

$\sqrt[\textcolor{red}{n}]{x} = {x}^{\frac{1}{\textcolor{red}{n}}}$

$4 \sqrt[\textcolor{red}{8}]{4} = 4 \cdot {4}^{\frac{1}{\textcolor{red}{8}}}$

We can now use these two rules of exponents to again rewrite this expression:

$a = {a}^{\textcolor{red}{1}}$ and ${x}^{\textcolor{red}{a}} \times {x}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} + \textcolor{b l u e}{b}}$

$4 \cdot {4}^{\frac{1}{8}} \implies {4}^{\textcolor{red}{1}} \times {4}^{\textcolor{b l u e}{\frac{1}{8}}} = {4}^{\textcolor{red}{1} + \textcolor{b l u e}{\frac{1}{8}}} = {4}^{\textcolor{red}{\frac{8}{8}} + \textcolor{b l u e}{\frac{1}{8}}} = {4}^{\frac{9}{8}}$

Or, depending on the solution you are looking for, using the reverse of the rule above:

${4}^{\frac{9}{8}} = {4}^{9 \cdot \frac{1}{8}} = {\left({4}^{9}\right)}^{\frac{1}{8}} \implies \sqrt[8]{{4}^{9}}$

Some of the simplifications are, depending on what you are working on:

$4 \sqrt[8]{4}$ or ${4}^{\frac{9}{8}}$ or $\sqrt[8]{{4}^{9}}$