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!