Show that$ f(x)=x^5-3$ is solvable by radicals over $\mathbb{Q}$.

Question

I was reading about solvability of quintics by radicals, but unfortunately there were no many examples and I am afraid that I do not understand the whole concept. How to show $x^5-3$ is solvable by radicals over $\mathbb{Q}$?

Answer 1

The splitting field $L =\mathbb{Q}[x]/(x^5-3)$ has 20 elements, and the Galois Group is

$$\text{Gal}(L\backslash\mathbb{Q}) \cong \langle (1,2)(3,4),(2,3,4,5)\rangle$$

Addition:

You can find the roots by using de Moivre's theorem. They are $\sqrt[5]{3}(\cos(\frac{2\pi n}{5})+\operatorname{i}\sin(\frac{2\pi n}{5})).$

Well-known formulae tell us that $\sin(\frac{\pi}{5}) = \frac{1}{4}(\sqrt{10-2\sqrt{5}})$ and $\cos(\frac{\pi}{5}) = \frac{1}{4}(1+\sqrt{5}).$

Using the formula $\sin(\alpha \pm \beta) = \sin\alpha\cos\beta \pm \sin\beta\cos\alpha$ and $\cos(\alpha\pm\beta) = \cos\alpha\cos\beta \mp \sin\alpha\sin\beta$ will get you closed form expressions for all of the roots.