radical extensions over $\mathbb{Q}$ of containing $(\sqrt{6} + 2 \sqrt[3](5))^4$

Question

I want to come up with a sequence of radical extensions for $(\sqrt{6} + 2 \sqrt[3]{5})^4$ over $\mathbb{Q}$. I started with letting $$F_0 = \mathbb{Q}.$$ I wouldn't have a problem if this were not raised to 4 since I would just have: $$\text{ let } a=\sqrt{6}+2\sqrt[3]{5}$$ then, $$F_1=F_0[x^2=6]=\mathbb{Q}(\sqrt{6}),$$ $$F_2 = F_1[x^3=8*5=40]=\mathbb{Q}(\sqrt{6},\sqrt[3]{5}).$$ Thus, I have $$\mathbb{Q}\subseteq \mathbb{Q}(\sqrt{6})\subseteq \mathbb{Q}(\sqrt{6},\sqrt[3]{5}).$$ I hope I am right with this. Now, when I try to get the sequence of radical extensions containing $$a^4=(\sqrt{6}+2\sqrt[3]{5})^4,$$ I become confused with what to do. Any help would be much appreciated. Thank you.

Answer 1

Well, let’s call your funny number $\alpha$, for convenience’s sake. You have $\Bbb Q\subset\Bbb Q(\alpha)\subset\Bbb Q\bigl(\sqrt6,\sqrt[3]5\,\bigr)=K$, and $[K:\Bbb Q]=6$. I don’t have time to check, but it seems supremely unlikely to me that $\alpha$ would be either quadratic or cubic over $\Bbb Q$. So adjoin $\sqrt6$ first, then $\sqrt[3]5$.