# How do you rationalize the denominator and simplify [1 / {1 + x^(1/6)}]?

##### 1 Answer
Jun 20, 2017

We need to multiply the numerator and the denominator by a polynomial, $P \left(x\right)$, such that the denominator becomes ${1}^{6} - {\left({x}^{\frac{1}{6}}\right)}^{6}$

#### Explanation:

Please understand that ${\left({x}^{\frac{1}{6}}\right)}^{6} = {x}^{\frac{6}{6}} = {x}^{1} = x$ and the denominator will be rational.

Let $u = {x}^{\frac{1}{6}}$, then #x = u^6

The denominator that we want can be written in terms of u as:

$- {u}^{6} + 1$

The denominator that we have can be written in terms of u as:

$u + 1$

We can find $P \left(u\right)$ by dividing the denominator that we want by the denominator that we have (please observe that I have filled in the missing terms with 0s):

$\textcolor{w h i t e}{\frac{u + 1}{\textcolor{b l a c k}{u + 1}}} \frac{- {u}^{5} + {u}^{4} - {u}^{3} + {u}^{2} - u + 1 \textcolor{w h i t e}{. . + 1}}{\text{)} \textcolor{w h i t e}{x} - {u}^{6} + 0 {u}^{5} + 0 {u}^{4} + 0 {u}^{3} + 0 {u}^{2} + 0 u + 1}$

$P \left(u\right) = - {u}^{5} + {u}^{4} - {u}^{3} + {u}^{2} - u + 1$

Reverse the substitution:

$P \left(x\right) = - {x}^{\frac{5}{6}} + {x}^{\frac{4}{6}} - {x}^{\frac{3}{6}} + {x}^{\frac{2}{6}} - {x}^{\frac{1}{6}} + 1$

We multiply the given expression by 1 in the form of $\frac{P \left(x\right)}{P \left(x\right)}$

$\frac{1}{1 + {x}^{\frac{1}{6}}} \frac{- {x}^{\frac{5}{6}} + {x}^{\frac{4}{6}} - {x}^{\frac{3}{6}} + {x}^{\frac{2}{6}} - {x}^{\frac{1}{6}} + 1}{- {x}^{\frac{5}{6}} + {x}^{\frac{4}{6}} - {x}^{\frac{3}{6}} + {x}^{\frac{2}{6}} - {x}^{\frac{1}{6}} + 1} =$

$\frac{- {x}^{\frac{5}{6}} + {x}^{\frac{4}{6}} - {x}^{\frac{3}{6}} + {x}^{\frac{2}{6}} - {x}^{\frac{1}{6}} + 1}{1 - x}$

This has rationalized the denominator.