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my question says:

The polynomials $x^2 − 5$ and $x^3 − 5$ are irreducible over $\mathbb Q$ (since they have no rational roots and their degrees do not exceed 3).

Determine the minimal polynomials for $\sqrt 5$ and $\sqrt[3]5$ over $\mathbb Q$, and determine the degrees $[\mathbb Q(\sqrt 5) : \mathbb Q]$ and $[\mathbb Q(\sqrt[3]5) : \mathbb Q]$.

Have they not already given me the minimal polynomials $x^2-5$ and $x^3-5$ giving degrees of 2 and 3 over $\mathbb Q$ respectively?

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Yes, they actually had (implicitly) given you that. Now compute the degrees knowing that e.g. the degree of $Q(\sqrt 5)$ over $Q$ is equal to the degree of the minimal polynomial of $\sqrt 5$ over $Q$.

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  • $\begingroup$ Huh that's strange they gave it to me like that. Also have I not done the degree computing already with [Q(√5):Q]=2 and [Q(cuberoot((5):Q]=3 or have I missed something? Thanks for helping. $\endgroup$ – Harumph Mar 21 '17 at 18:06

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