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Let rational function field $K=\mathbb{R}(X)$ , $F= \mathbb{R}(X^4 -\frac{1}{X^4 })$ and $L$ be the Galois closure of $K/F$.

I can't write L and don't understand $[L:F]$ .

Please find $[L:F]$.

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$K = F(X)$, so the Galois closure of $K$ over $F$ is the splitting field of the minimum polynomial of $X$ over $F$.

The minimum polynomial of $X$ over $F$ is $$t^8 - (X^4 - \frac 1 {X^4})t^4 - 1.$$

This has roots $\pm X, \pm iX, \pm \frac {\zeta_8} X, \pm \frac {i\zeta_8} X$. So the Galois closure is $\mathbb C(X)$.

Does this help?

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    What is the reason why the polynomial written as the minimum polynimial of $X$ over $F$ is irreducible? – 2017-02-27
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    It is the polynomial of lowest degree in $F[t]$ that has $X$ as a root. You can't find a polynomial of any lower degree. – 2017-02-27
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    I see. Thank you so much, Kenny Wong!! – 2017-02-27
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    @KennyWong To see that the polynomial is minimal, it might help to consider the chain of extensions $\mathbb R\left(X^4-\frac{1}{X^4}\right) \subset \mathbb R(X^4) \subset \mathbb R(X)$. The latter extension can readily be seen to have degree $4$, and the former has degree $2$, so the composite extension has degree $8$. – 2017-02-27
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    @DustanLevenstein Good point - that's faster than my brute force approach! – 2017-02-27
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    Oops, sorry, I meant to tag @loglog. – 2017-02-27