In most of the text books in algebra describes the splitting field of $x^3-2$ over $\mathbb{Q}$ as $\mathbb{Q}(2^{\frac{1}{3}}, \rho)$ where $\rho$ is the primitive root of unity ($\rho=e^{\frac{i2\pi}{3}})$.
The splitting field is also equal to $\mathbb{Q}(2^{\frac{1}{3}}, i\sqrt 3)$. Can we use this generators of the splitting filed and compute the Galois group?
For Example, an automorphism will map $2^{\frac{1}{3}}$ to the roots of the minimal polynomial of $2^{\frac{1}{3}}$ which is $x^3-2$ (and the roots are $2^{\frac{1}{3}}, 2^{\frac{1}{3}}\rho , 2^{\frac{1}{3}}\rho ^2$)
And an automorphism will map $i\sqrt 3$ to the roots of the minimal polynomial of $i\sqrt 3$ which is $x^2+3$ (and the roots are $i\sqrt 3, -i\sqrt 3$).
Hence, there will be total $6$ automorphisms.
My motivation behind choosing these generators are to avoid any relations between generators and to avoid invalid generators.
My question is to valid to choose these generators $2^{\frac{1}{3}}, i\sqrt 3$?