# How do you rationalize the denominator and simplify root3( (1 / (2x^2)))?

Aug 20, 2017

You $\underline{\text{force}}$ it into something you can take the cube root of.

$= \frac{\sqrt[3]{4 x}}{2 x}$

#### Explanation:

Multiply by 1 and you do not change the intrinsic value. However, 1 comes in many forms

$\textcolor{g r e e n}{\sqrt[3]{\frac{1}{2 {x}^{2}} \textcolor{red}{\times 1}}}$

$2 \times {2}^{2} = {2}^{3}$

${x}^{2} \times {x}^{2} \times {x}^{2} = {x}^{2} \times {\left({x}^{2}\right)}^{2} = {\left({x}^{2}\right)}^{3}$

$\textcolor{g r e e n}{\sqrt[3]{\frac{1}{2 {x}^{2}} \textcolor{red}{\times \frac{{2}^{2} {\left({x}^{2}\right)}^{2}}{{2}^{2} {\left({x}^{2}\right)}^{2}}}}} \text{ "->" } \sqrt[3]{\frac{4 {x}^{4}}{{2}^{3} {\left({x}^{2}\right)}^{3}}}$

color(white)("vvvvvvvvvvvvvvv")->" "root(3)(4x^4)/(2x^2)
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Simplified further

${x}^{4} \to {x}^{3} \times x \text{ }$ so we can change the $4 {x}^{4}$ such that we have:

$\frac{x \sqrt[3]{4} x}{2 {x}^{2}} = \frac{x}{x} \times \frac{\sqrt[3]{4 x}}{2 x}$

$= \frac{\sqrt[3]{4 x}}{2 x}$