I don't really understand the notion of a radical extension, and have been struggling to figure out how to show that $\mathbb{Q} \subset K $ is a radical extension when $K = \mathbb{Q}(\sqrt[5]{1+\sqrt{3}})$. My best guess is that this can be done by constructing the tower $\mathbb{Q} \subset \mathbb{Q}(\sqrt{3}) \subset \mathbb{Q}(\sqrt{3}, \sqrt[5]{1+\sqrt{3}})$.
Any advice would be apprciated!
Yes, it is a radical extension by definition. And also the Galois group is given by the Galois group $G$ of the minimal polynomial $f=x^{10}-2x^5-2$ of $\alpha=\sqrt[5]{1+\sqrt{3}}$. It is known that $G\cong F_5\times C_2$ is a solvable group of order $200$. Hence $K/\Bbb Q$ is solvable by radicals.