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(a) Is $\sqrt{1+\sqrt[3]{2}}$ is constructible?

(b)Is $1+\sqrt{2}+\sqrt[4]{3}$ constructible?

Attempt:

(a) Let $X=\sqrt{1+\sqrt[3]{2}}$. Then, upon squaring, we have $(X-1)^2=\sqrt[3]{2}$. Cubing, we get $X^6-3X^4+3X^2-3=0$. By Eisenstein with $p=3$, we have that this polynomial is irreducible, and so it is indeed the minimum polynomial of $\sqrt{1+\sqrt[3]{2}}$. Since the minimum polynomial has degree larger than $5$, it is not constructible.

(b) I think we could do the same method as above, but it would be quite tedious (lots of squaring), so there must be a better method. My very rough idea would be that as $\sqrt{2}\not\in\mathbb{Q}(\sqrt[4]{3})$ we have $[\mathbb{Q}(1+\sqrt{2}+\sqrt[4]{3}):\mathbb{Q}]=[\mathbb{Q}(\sqrt{2},\sqrt[4]{3}):\mathbb{Q}]=[\mathbb{Q}(\sqrt{2}:\sqrt[4]{3})][\mathbb{Q}(\sqrt[4]{3}):\mathbb{Q}]=2\times 4=8$, and this is the degree extension (which in turn is equal to the degree of the minimum polynomial), and therefore as the degree is larger than $5$, this number is not constructible.

Can anybody comment on my solutions? Are they correct?

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    No they are not. The condition for an algebraic number to be constructible is not that its degree is $\le 5.$. – Anne Bauval Dec 17 '22 at 14:51
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    For $(b)$ especially, remember that $\mathbb{K}$ the set of real constructible numbers is a field, and on which $\sqrt{x} \in \mathbb{K}$ for all $x \in \mathbb{K}$. As for the condition for being constructible, see the bullet point 1 of this question. – Bruno B Dec 17 '22 at 14:59
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    @AnneBauval oops yeah I remembered that the degree of the extension must be a power of $2$ (necessary but insufficient). So in part (a) we indeed have non constructibility right? But I'm unsure how to proceed with (b) –  Dec 17 '22 at 15:01
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    Ok for (a). And (b) is much simpler: follow Bruno B's hint. – Anne Bauval Dec 17 '22 at 15:02
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    I forgot to put $x \geq 0$, my bad (and I wish you could edit comments after the five minutes limit) – Bruno B Dec 17 '22 at 15:34

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