Q:

Is $\mathbb{Q}(\sqrt{2}, 2^{1/3})$ a Galois extension of $\mathbb{Q}$?

I am wondering about the following question: 
Is $\mathbb{Q}(\sqrt{2}, 2^{1/3})$ a Galois extension of $\mathbb{Q}$ ? 
For a finite extension $E/F$, the statement "$E/F$ is Galois" is equivalent to the statement "$E$ is a splitting field of a separable polynomial with coefficients in $F$". 
Per the above statement, I believe that $\mathbb{Q}(\sqrt{2}, 2^{1/3})$ is not a Galois extension of $\mathbb{Q}$. This is because, it doesn't seem to me that $\mathbb{Q}(\sqrt{2}, 2^{1/3})$ is even a splitting field of any polynomial with coefficients in $\mathbb{Q}$. 
Am I correct? If so, how can I make that last sentence above more rigorous, and actually show it's not a splitting field of any polynomial with coefficients in $\mathbb{Q}$ ? 
Thanks!

A:

Hint:  It isn't a normal extension, as it doesn't have all the roots of $x^3-2$.

