Root in a splitting field

Question

For this question, where does $(-3^{1/3}+-(3^{1/6})^5i)/2$ come from?

Answer 1

If you solve $x^3-3$, one solutions you get is $\sqrt[3] 3$. To get the other solutions, first, factor out $x-\sqrt[3] 3$ from the polynomial:

$$x^3-3=x^3-\sqrt[3] 3 x^2+\sqrt[3] 3 x^2-(\sqrt[3] 3)^2 x+(\sqrt[3] 3)^2 x-3=(x-\sqrt[3] 3)(x^2+\sqrt[3] 3 x+(\sqrt[3] 3)^2)$$

Now, for the quadratic, using the quadratic formula, we get:

$$\frac{-\sqrt[3] 3 \pm \sqrt{(\sqrt[3] 3)^2-4(\sqrt[3] 3)^2}}{2}=\frac{-\sqrt[3] 3 \pm \sqrt{-3(\sqrt[3] 3)^2}}{2}=\frac{-\sqrt[3] 3 \pm (\sqrt[6] 3)^5 i}{2}$$