How to simplify this cube root in denominator expression?

Question

I have the expression:

$$\large\frac{1}{x^2 + \sqrt[3]{2x} + \sqrt[3]4}$$

I'm not sure how to simplify this, because it seems difficult to remove the radical from the denominator.

I know $\sqrt[3]4$ reduces to $\sqrt[3]2 \cdot \sqrt[3]2$ but that doesn't seem to get me too far.

Thanks

Answer 1

$$ \begin{gathered} \frac{1} {{\left( {x^2 + \sqrt[3]{2}x + \sqrt[3]{4}} \right)}} = \frac{{\left( {x - \sqrt[3]{2}} \right)}} {{\left( {x^2 + \sqrt[3]{2}x + \sqrt[3]{4}} \right)\left( {x - \sqrt[3]{2}} \right)}} = \hfill \\ \hfill \\ = \frac{{\left( {x - \sqrt[3]{2}} \right)}} {{x^3 - 2}} \hfill \\ \end{gathered} $$

using $$(a^3-b^3)=(a-b)(a^2+ab+b^2)$$