Proving that the unit cube cannot be tripled (with straight edge and compass) 
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)?

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.
