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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

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  • $\begingroup$ Your expression is not clear. When using \sqrt[3]{...}, remember to put what should be inside the root in the place of the ... Also, check the parenthesis. $\endgroup$ – Daniel Sep 16 at 19:20
  • $\begingroup$ Ah yes, my bad, fixed it! $\endgroup$ – Evan Sep 16 at 19:21
  • $\begingroup$ It is still not clear whether it is $(\sqrt[3]{2})x$ leading to nice Luca's solution or $2x$ inside the cube root (as you wrote) which does not simplify so well. $\endgroup$ – zwim Sep 16 at 19:40
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$$ \begin{gathered} \frac{1} {{\left( {x^2 + \sqrt[3]{2}x + \sqrt[3]{4}} \right)}} = \frac{{\left( {x - \sqrt[3]{2}} \right)}} {{\left( {x^2 + \sqrt[3]{2}x + \sqrt[3]{4}} \right)\left( {x - \sqrt[3]{2}} \right)}} = \hfill \\ \hfill \\ = \frac{{\left( {x - \sqrt[3]{2}} \right)}} {{x^3 - 2}} \hfill \\ \end{gathered} $$

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

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  • $\begingroup$ This is probably right... More precisely, it is correct if instead of $\sqrt[3]{2x}$ the OP actually meant $\sqrt[3]{2} x$. $\endgroup$ – Daniel Sep 16 at 19:25
  • $\begingroup$ Yep, that's what I mean, thanks! $\endgroup$ – Evan Sep 16 at 19:25
  • $\begingroup$ This is the right form. Sorry for my typos $\endgroup$ – Luca Sep 16 at 19:31
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    $\begingroup$ resulting formula is not homogeneous in degree, denominator should be in $x^3$. $\endgroup$ – zwim Sep 16 at 19:37
  • $\begingroup$ Yes, of course! Sorry again. I'm very tired and I get a lot of typos. Better I stop here. Sorry again $\endgroup$ – Luca Sep 16 at 19:40

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