How can I show that $\mathbb{Q}(\alpha^3)\subsetneq\mathbb{Q}(\alpha^3\sqrt{2})$? I am struggling with this field extensions problem.

Let $\alpha \in \mathbb{C}$ be a root of $x^5+7x^2-14x+14 \in \mathbb{Q}[X]$. Show whether the following statement is true or false:
$$
\mathbb{Q}(\alpha) = \mathbb{Q}(\alpha^3) \subsetneq \mathbb{Q}(\alpha^3\cdot\sqrt{2})
$$

I have already proved that $\mathbb{Q}(\alpha) = \mathbb{Q}(\alpha^3)$, but I'm not able to show the second part so I'm not even sure whether it's true or false. I think that I must use the field $\mathbb{Q}(\alpha^3,\sqrt{2})$ whose degree over $\mathbb{Q}$ is 10, but I can't conclude anything.
Any suggestions? Thanks for your helping.
 A: You have already established that $\beta=\alpha^3$ is of degree $5$ over rationals and hence we can conclude that the minimal polynomial of $\beta$ is of the form $$x^5+ax^4+bx^3+cx^2+dx+e\in\mathbb {Q } [x] $$ and in particular we have $e=14^3\neq 0$.
Let $\gamma=\beta\sqrt{2}$ so that $\beta=\gamma/\sqrt{2}$ is a root of above polynomial. And thus we get $$\gamma^5+a\gamma^4\sqrt{2}+2b\gamma^3+2c\gamma^2\sqrt{2}+4d\gamma+4e\sqrt{2}=0$$ or $$\gamma^5+2b\gamma^3+4d\gamma+(a\gamma^4+2c\gamma^2+4e)\sqrt{2}=0\tag{1}$$ Next we observe that $$a\gamma^4+2c\gamma^2+4e=4(a\beta^4+c\beta^2+e)$$ and the above can't be zero as $\beta$ is of degree $5$ over $\mathbb {Q} $. Hence from $(1)$ we get $$\sqrt{2}=-\frac{\gamma^5+2b\gamma^3+4d\gamma}{a\gamma^4+2c\gamma^2+4e}$$ and thus $\sqrt{2}\in\mathbb {Q} (\gamma) $. Further $\beta=\gamma/\sqrt{2}$ and $\beta\in\mathbb {Q} (\gamma) $ as well. Thus we have $\mathbb{Q} (\beta, \sqrt{2})\subseteq \mathbb {Q} (\gamma) $. The inclusion $\mathbb {Q} (\gamma) \subseteq \mathbb {Q} (\beta, \sqrt{2})$ follows from the fact that $\gamma=\beta\sqrt{2}$. Thus we have $\mathbb{Q} (\beta\sqrt{2})=\mathbb{Q} (\beta, \sqrt{2})$ and this is of degree $10$ over $\mathbb {Q} $. Thus we have $\mathbb {Q} (\alpha ^3)\subsetneq\mathbb {Q} (\alpha^3\sqrt{2})$.
