$[\mathbb{Q}(2^{1/4}+8^{1/2}):\mathbb{Q}]=\text{degree(minimal polynomial)}$
I think that the minimal polynomial is $(x^4+48x^2+62)^2=2(8x^3-64x)^2$ with degree $ 8$, but it's not irreducible by Eisenstein's theorem.
any suggest, for resolution?
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$\begingroup$ I don't think this is a degree 8 extension as $\sqrt{8}=2\sqrt{2}$ and this is already in $\mathbb{Q}(2^{\frac{1}{4}})$. $\endgroup$– Sam WeatherhogNov 9, 2015 at 22:00
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$\begingroup$ One does not use Eisenstein to prove non-irreducibility. As a simple example, $x^2+4$ "fails" Eisenstein, or more properly Eisenstein does not apply, but $x2+4$ is irreducible over the rationals. $\endgroup$– André NicolasNov 9, 2015 at 22:08
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1 Answer
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Since $\sqrt{8}=2\sqrt{2}$, once you add $\sqrt[4]{2}$, you automatically added its square ($\sqrt{2}$) and so also added its square doubled ($2\sqrt{2}$). Hence you are only adding $\sqrt[4]{2}$.
