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I am asked to find the degree and basis for a given field extension $\mathbb{Q}(\sqrt[3]{2},\sqrt[3]{6},\sqrt[3]{24}) $

Now I know that the degree for each vector is $3$, and that the basis will have $9$ vectors. I found the answer in the back of the book as $\{1,\sqrt[3]{ 2},\sqrt[3]{ 4},\sqrt[3]{ 3},\sqrt[3]{ 6},\sqrt[3]{ 12},\sqrt[3]{9},\sqrt[3]{ 18},\sqrt[3]{ 36}\}$ but I would like to know how you find them. Thanks!

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    How do you know the basis will have 9 vectors?2012-02-22
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    Chris is right: this is not trivial at all! Equivalently, it is a real problem to show that $[\mathbb Q( \sqrt[3]{2},\sqrt[3]{3}):\mathbb Q]=9$2012-02-22

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