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I would like to show the unit cube cannot be tripled using a straightedge and compass.

I note that the side of a cube that has been tripled would have a side length of $\alpha=\sqrt[3]{3}$

But then the minimal polynomial for this $\alpha$ is $f(x)=x^3-3$

Then $[\mathbb{Q}(\alpha):\mathbb{Q}]=3\neq 2^{n}$ for $n\in\mathbb{N}$ and hence our choice of $\alpha=\sqrt[3]{3}$ cannot be constructible and hence we cannot triple the cube.

This is relevant because I know that $\alpha$ is constructible only if $[\mathbb{Q}(\alpha):\mathbb{Q}]=2^n$ for some $m\in\mathbb{N}$

I was wondering if this approach to the proof was okay, or would anyone recommend proving it differently/better or correctly (if this approach is incorrect)?

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  • $\begingroup$ Your proof is good, but I think there are elements with power-of-two degree which are not constructible. $\alpha$ is constructible iff it is in a radical tower over $\mathbb{Q}$, which implies $[\mathbb{Q}(\alpha) : \mathbb{Q}] = 2^n$, but the reverse implication needn't hold. $\endgroup$ – TorsionSquid Apr 19 '15 at 2:12
  • $\begingroup$ Yes I think you're right. I wasn't sure if this was the correct mode of attack for this problem since wouldn't this means regardless if I was tripling it or quadrupling it, etc, then dimension of $\mathbb{Q}(\alpha)$ over $\mathbb{Q}$ would always be $3$ and hence the unit cube would never be able to tripled, quadrupled, etc...? $\endgroup$ – James Apr 19 '15 at 2:18
  • $\begingroup$ No I think your solution is fine. I think it does follow in a similar way that the cube cannot be quadrupled. But, for example, this method would not work to show that the cube cannot be duplicated 27-fold, since $\sqrt[3]{27}$ is not degree three. $\endgroup$ – TorsionSquid Apr 19 '15 at 2:21
  • $\begingroup$ Ah, yes, I see what you mean. Well, you've answered my question, so if you want to change the comment to an answer I'll be happy to choose you as accepted answer. $\endgroup$ – James Apr 19 '15 at 2:23
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Your solution is fine. It does follow in a similar way that the cube cannot be quadrupled. But, for example, this method would not work to show that the cube cannot be duplicated 8-fold, since $\sqrt[3]{8}$ is not degree three.

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