Infinite-dimensional extensions of $\mathbb Q$ I need help to solve the following exercise:

Let $X$ be an indeterminate over $\mathbb Q$ (so a transcendental number) and consider the field extensions $\mathbb Q\subseteq \mathbb Q(X^3)\subseteq\mathbb Q(X^2)\subseteq\mathbb Q(X)$. Prove that $$\mathrm{Fix}(\mathrm{Gal}(\mathbb Q(X)/\mathbb Q(X^2)) )= \mathbb Q(X^2)$$ and that $$\mathrm{Fix}(\mathrm{Gal}(\mathbb Q(X)/\mathbb Q(X^3)) )\supsetneq\ \mathbb Q(X^3).$$

I hope that the notations $\mathrm{Gal}({}\cdot{}/{}\cdot{})$ and $\mathrm{Fix}({}\cdot{})$ are quite standard and so understandable.
My considerations: The subgroups $H\subseteq G=\mathrm{Gal}(\mathbb Q(X)/\mathbb Q)$ that satisfy the condition $\mathrm{Gal}(G/\mathrm{Fix}(H))=H$ are only the finite subgroups of $G$. So one way to solve the problem could be showing that $\mathrm{Gal}(\mathbb Q(X)/\mathbb Q(X^2))$ is a finite group, but $\mathrm{Gal}(\mathbb Q(X)/\mathbb Q(X^3))$ is an infinite group.
Thanks in advance
 A: I'd rather write $\;t\;$ all through instead of $\,x\,$ , which can be misleading (for me, in particular) for the variable/unknown/indeterminate used for functions and/or polynomials. 
Now, we can write
$$\Bbb Q(t)=\Bbb Q(t^2)[t]$$
since the $\,t\,$ is algebraic over $\,\Bbb Q(t^2)\,$ as it is a root of the quadratic $\,f(X):=X^2-t^2\in\Bbb Q(t^2)[X]\,$ (remember that if $\,F/k\,$ is a fields extension and $\,w\in F\;$ , then $\,w\,$ is algebraic over $\,k\,$ iif $\,k(w)=k[w]\,$
Since clearly both $\,t\,,\,-t\in\Bbb Q(t)\,$ , the extension is normal (and algebraic and separable) and thus Galois, and we're done with the first task
OTOH, we also have that $\,\Bbb Q(t)=\Bbb Q(t^3)[t]\,$ for the same reason as above, with $\,g(X)=X^3-t^3\in\Bbb Q(t^3)[X]\;$ , yet this time we get that
$$X^3-t^3=(X-t)(X^2+tX+t^2)$$
and the above quadratic's discriminant is
$$\Delta=t^2-4t^2=-3t^2\;,\;\;\text{and}\;\;\sqrt{-3t^2}\notin\Bbb Q(t^3)\;\text{(why?)}$$
Thus, the extension is this case is not normal and thus not Galois, and in fact $\;Gal(\Bbb Q(t)/\Bbb Q(t^3))=1\;$ , and this group's fixed field is...
