It is easy to see that, $[\Bbb{Q}(\sqrt{3},\sqrt[3]2)]=3$. Therefore the dimension of the vector space $\Bbb{Q}(\sqrt{3},\sqrt[3]{2})$ over $\Bbb{Q}(\sqrt{3})$ is 3. Thus basis will contain $3$ elements.
How to find the basis of this problem? Any help will be appreciated.
Thanks!
If we find the minimal polynomial of $\sqrt[3]{2}$ over the $\Bbb Q (\sqrt{3})$ then degree of this polynomial is the dimension of the vector space $\Bbb Q (\sqrt[3]{2},\sqrt{3})/ \Bbb Q(\sqrt{3})$ and the powers of $\sqrt[3]{2}$ up to degree of the minimal polynomial are a basis of that extension. Now the polynomial $x^{3}-2$ is an irreducible polynomial over $\Bbb Q(\sqrt{3})$. So the elements $1 , \sqrt[3]{2}, (\sqrt[3]{2})^2$ are a basis for this space.