Let $E=\mathbb{Q}(2^{1/3})$. What is the normal closure of $E/E$? 
Let $E=\mathbb{Q}(2^{1/3})$. What is the normal closure of $E/E$?

My thought is the $A(2^{1/3})$ where $A$ is an algebraic closure of $\mathbb{Q}$. But I am not sure whether it is correct and why...
Thanks for your time.
 A: The field $\;A(\sqrt[3]2)\;$, with $\;A=\overline{\Bbb Q}\;$ is way, way too big for being the normal closure of the given field. You only need to "enlarge" the given field the exact amount in order to make the extension "normal" without losing algebraicity.
Now, putting $\;\alpha:=\sqrt[3]2\;$, we see the minimal polynomial of this element over the rationals is $\;x^3-2\;$ , and since all the roots of this pol. are  $\;\alpha,\,\alpha w,\,\alpha w^2\;,\;\;w:=e^{2\pi i/3}\;$,  the extension $\;\Bbb Q(\alpha,w)/\Bbb Q\;$ already contains $\;\Bbb Q(\alpha)/\Bbb Q\;$ and it is already normal as it is the splitting field of $\;x^3-2\;$ over $\;\Bbb Q\;$.
Thus, you've already found your normal closure: it is $\;\Bbb Q(\alpha,w)/\Bbb Q\;$ .
A: The normal closure of $E/E$ is $E$. 
Note that a normal closure of a finite extension is always a finite extension, excluding your answer. Also note that $A[\sqrt[3]{2}]$ is in fact equal to $A$. 
Recall that a characterization of normal is: "Every irreducible polynomial in K[X] that has one root in L, has all of its roots in L, that is, it decomposes into linear factors in L[X]."
Yet the only way that a polynomial $f$ in $E[X]$ that is irreducible can have a root in $E$ is that it is of degree $1$. 
Note that the polynomial $x^3 -2$ is not irreducible over $E$. Therefore it is no longer a problem that it does not decompose.
Or, for example the polynomial $x^2 - 2$ has no root in $E$, so it is not relevant it does not decompose. 
