Is it possible to express $\sqrt[6]{2}$ as rational combination of $1, \sqrt{2}, \sqrt[4]{2}$? Is it possible to express $\sqrt[6]{2}$ as rational combination of $1, \sqrt{2}, \sqrt[4]{2}$?
So can we find $a,b,c \in \mathbb{Q}$ such that $\sqrt[6]{2}=a+ b\sqrt{2}+c \sqrt[4]{2}$?
My guess is no and one should try to prove by contradiction. And the contradiction should show that either at least one number from ${a, b, c}$ is irrational or at least one number from ${\sqrt{2}, \sqrt[4]{2}}$ is rational.
However, my attempts failed because they lead to some unmanageable (for me at least) long computations. 
Can you please prove some smart way to handle this?
Thanks a lot!
 A: Hint: By Eisenstein's Criterion, $x^{12}-2$ is irreducible over $\mathbb{Q}$. Therefore, there are no $a,b,c,d\in\mathbb{Z}$ so that $ax^6+bx^3+cx^2+d=0$ where $x=2^{1/12}$.
A: By Eisenstein's Criterion, you know that $x^6-2$ is irreducible over $\mathbb{Q}$.  Observe that $\sqrt[6]{2}$ is a root of this polynomial.
Suppose (for contradiction) that $\sqrt[6]{2}=a+b\sqrt{2}+c\sqrt[4]{2}$ for some rational $a$, $b$, and $c$.  Let $y=a+b\sqrt{2}+c\sqrt[4]{2}$.  If you look at $y^0,\cdots,y^4$ as vectors in the span of $1$, $\sqrt[4]{2}$, $\sqrt{2}$, and $\sqrt[4]{8}$, then this is a collection of $5$ vectors in a vector space of dimension at most $4$, so there is some nontrivial combination of these which equals zero.  This tells you that there is a nonzero polynomial $p$ of degree at most $4$ with rational coefficients with $y$ as a root (since $y^0,\dots,y^4$ is a linear dependent set).
Therefore, $\sqrt[6]{2}$ is a root of both $x^6-2$ and $p$, so it is a root of their gcd (which cannot be $1$ because the gcd needs a root).  This, however, is impossible because such a gcd would need to be of degree at most $4$ and at least $1$ and divide $x^6-2$, but $x^6-2$ is irreducible.  This is a contradiction.
