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The question reads: Find a basis of $\mathbb{Q}(\sqrt{2}+\sqrt[3]{4})$ over $\mathbb{Q}$. Describe the elements of $\mathbb{Q}(\sqrt{2}+\sqrt[3]{4})$.

My original thought on the approach was to find the minimum polynomial (which would have degree 6), and then just take create a basis where every element is $(\sqrt{2}+\sqrt[3]{4})^n$ for $n=0,\ldots,5$.

The hint in the back recommends adjoining $\sqrt[3]{4}$ first, but I'm not sure where to go once I've done that. Can anyone help guide me through the process the book wants me to use?

Thank you very much in advance.

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    While the process suggested by the book and fleshed out by Arturo is probably the nicest way to solve the problem, there is nothing wrong with your original thought. Find a polynomial of degree 6 satisfied by $\sqrt2+\root3\of4$, prove that it's irreducible, then those powers you wrote down form a basis.2012-04-14
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    There is another question which deals with exactly the same field extensions $\mathbb Q(\sqrt2+\sqrt[3]4)$: [Finding a basis for the field extension ${\mathbb{Q}(\sqrt{2}+\sqrt[3]{4})}$](http://math.stackexchange.com/questions/124425/finding-a-basis-for-the-field-extension-mathbbq-sqrt2-sqrt34/)2012-05-08

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