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.

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. 
