<s>[INST] Minimal polynomial of  $\sqrt[3]{7-\sqrt{2}}$ I want to find the minimal polynomial (over $\mathbb{Q}$) of: $k:=\sqrt[3]{7-\sqrt{2}}$.
With simple 'tricks' I got that: $P=(x^3-7)^2+2$ is a polynomial such that $P(k)=0$.
But I don't know if, or how to prove that $P$ is the minimal polynomial. How can I prove this/find the minimal polynomial ? 
 [/INST] Let $\alpha = \sqrt[3]{7 - \sqrt{2}}$ and $P(x) = x^6 - 14x^3 + 47$. Since $P(\alpha) = 0$, in order to show $P$ is irreducible it suffices to show $[Q(\alpha):Q] = 6$. Since $7 - \alpha^3 = \sqrt{2}$, you have that $\sqrt{2} \in Q(\alpha)$ and thus 
$$[Q(\alpha):Q] =[Q(\alpha):Q(\sqrt{2})][Q(\sqrt{2}):Q] = 2[Q(\alpha):Q(\sqrt{2})]$$
Thus our goal is to show that $[Q(\alpha):Q(\sqrt{2})] = 3$. Since $\alpha$ satisfies the third-degree polynomial equation $Q(x) = x^3 - 7 + \sqrt{2} = 0$ over $Q(\sqrt{2})$, our task is equivalent to showing $Q(x)$ is irreducible over  $Q(\sqrt{2})$.
But if $Q(x)$ were not irreducible over  $Q(\sqrt{2})$, it would have a linear factor. Thus it would have a root $a + b\sqrt{2}$, with $a$ and $b$ rational. But the roots of $Q(x) = 0$ are the numbers $\alpha$, $e^{2\pi i \over 3}\alpha$, and $e^{4\pi i \over 3}\alpha$. Of these, only $\alpha$ is real, so we have
$$\alpha =  a + b\sqrt{2}$$
Cubing, we get
$$7 - \sqrt{2} = (a + b\sqrt{2})^3$$
Thus we must have
$$7 + \sqrt{2} = (a - b\sqrt{2})^3$$
Multiplying the above two equations together we get
$$47 = (a^2 - 2b^2)^3$$
Since $47$ is not the cube of a rational number, we have arrived at a contradiction and we are done.
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