Why is $\Bbb Q(\sqrt[3]{2},\sqrt{-3})$ the splitting field of $x^3-2$?

Question

I just don't see it. Too my mind, it would suffer to show that $i$ or $\sqrt 3$ are inside the extension, so that you can construct a primitive third root of unity.

Answer 1

Note that $$x^3-1 = (x-1)(x^2+x+1) = (x-1)(x-\omega)(x - \bar{\omega})$$ where $$\omega = \frac{-1 + \sqrt{-3}}{2}$$

So a primitive third root of the unity can obtained with $\sqrt{-3}$ and vice versa.

Answer 2

You can actually write primitive 3rd root of unity as $\frac{-1+\sqrt{-3}}{2}$. If we denote it by $\omega$, then we clearly see that $\sqrt[3]{2},\sqrt[3]{2}\omega,\sqrt[3]{2}\omega^2$ are all in your field.