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I've been having problems with the following proof:

Let $\alpha_1,\alpha_2,\alpha_3$ be real numbers such that $(\alpha_i)^2 \in Q$ for each $i$, and let $K=Q(\alpha_1,\alpha_2,\alpha_3).$ Show that the cube root of 2 is not contained in $K$.

I've tried a contradiction method, where I assume the cube root is contained, and then showing that you can't construct it with the alphas, and I've tried fiddling with the degrees of the extensions for a contradiction, but I can't get it to lead anywhere.

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The degree argument should work. Show that the degree of $K$ over the rationals is 1, 2, 4, or 8; deduce that $K$ can't contain a field of degree 3 over the rationals; deduce further that $K$ can't contain $\root3\of2$.

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    I may be a victim of tunnel vision, but I can’t imagine any other way of showing this.2012-11-01