Finding the minimal polynomial of $\sqrt[3]{7-\sqrt{3}}$ over $\mathbb Q.$ By simple algebraic means I got that $P(x):=(x^3-7)^2-3$ is a polynomial s.t $P(\alpha)=0$ where $\alpha = \sqrt[3]{7-\sqrt{3}}$.
I wish to show that $P$ is of minimal degree, is this proof ok ? 
Clearly $\mathbb{Q}(\sqrt3)$ is of degree 2 over $\mathbb{Q}$ so If I show that $F(x):=x^3-7+\sqrt3$ is irreducible over $\mathbb{Q}(\sqrt3)$ then the degree is 6 [Here I am missing an argument, not too sure why this will end things]
I know that $F(x)$ have exactly 3 roots, 2 are in $\mathbb{R}$ and one is not.
So I also have to prove that the two real roots are not in $\mathbb{Q}(\sqrt3)$...
Any ideas ?
 A: If $x^3-7+\sqrt{3}$ is irreducible over $\mathbb{Q}(\sqrt{3})$, then the degree of the extension $\mathbb{Q}(\sqrt{3},\sqrt[3]{7-\sqrt{3}})$ over $\mathbb{Q}(\sqrt{3})$ will be $3$. By Dedekind's Product Theorem, that will mean that
$$[\mathbb{Q}(\sqrt{3},\sqrt[3]{7-\sqrt{3}}):\mathbb{Q}] = [\mathbb{Q}(\sqrt{3},\sqrt[3]{7-\sqrt{3}}):\mathbb{Q}(\sqrt{3})][\mathbb{Q}(\sqrt{3}):\mathbb{Q}] = 3\times 2 = 6.$$
Finally, you observe that $\sqrt{3}\in\mathbb{Q}(\sqrt[3]{7-\sqrt{3}})$, so that $\mathbb{Q}(\sqrt{3},\sqrt[3]{7-\sqrt{3}}) = \mathbb{Q}(\sqrt[3]{7-\sqrt{3}})$, and you would be done. 
But you can show directly that your polynomial
$$(x^3-7)^2 -3 = x^6 - 14x^3 + 46$$
is irreducible, e.g., by Eisenstein's or Schönemann's Criterion.
A: One way to finish: the norm $N_{\mathbb{Q}(\sqrt{3})/\mathbb{Q}}(7+\sqrt{3})=7^2-3=46$; it is not a cube, so $7+\sqrt{3}$ is not a cube in $\mathbb{Q}(\sqrt{3})$.
A: Eisenstein's Criterion (with $p=2$) implies that $P=x^6-14 x^3+46$ is irreducible.
A: Hint $\rm\ (i + j\sqrt{3})^3\:=\ k + 3\:m\: \sqrt{3}\:\neq\: n \pm \sqrt{3}$ 
