For which rational coefficients is $a+b\alpha+c\alpha^2+d\alpha^3+e\alpha^4$ constructible, where $\alpha=3^{1/5}$? 
Let be $\alpha=3^{1/5}$, and be $\gamma$ the number
$$
\gamma=a+b\alpha+c\alpha^2+d\alpha^3+e\alpha^4
$$
with $a,b,c,d,e \in \mathbb{Q}$ . For which $a,b,c,d,e \in \mathbb{Q}$ is $\gamma$ constructible?

I know the field of constructible numbers is closed by square roots and the Wantzel theorem for constructibility, but I didn't get anywhere.
Any hint?
 A: The number $\gamma$ lives in the field $\Bbb Q(\alpha)$, which is a degree $5$ extension of $\Bbb Q$.
If $\gamma$ is constructible, then the extension $\Bbb Q(\gamma)/\Bbb Q$ must have power-of-two degree.
But this degree should also be a factor of $5$, as $\Bbb Q(\gamma)/\Bbb Q$ is a subextension of $\Bbb Q(\alpha)/\Bbb Q$.
Therefore the only possibility is that the degree is equal to $1$, i.e. $\gamma \in \Bbb Q$. This again is equivalent to $b = c = d = e = 0$, as $1, \alpha,\dots, \alpha^4$ form a $\Bbb Q$-basis of $\Bbb Q(\alpha)/\Bbb Q$.
A: Let me give a hint, using numbers having the form $a+b\sqrt[3]2+c\sqrt[3]4$, with $a,b,c\in\mathbb{Q}$.  Whatever irreducible polynomial with rational coefficients has this root, will have a set of three conjugate roots including the given one ($\omega=$ a non-real cube root of unity):
$r_1=a+b\sqrt[3]2+c\sqrt[3]4$
$r_2=a+b\omega\sqrt[3]2+c\omega^2\sqrt[3]4$
$r_3=a+b\omega^2\sqrt[3]2+c\omega\sqrt[3]4$
and these would also be constructible.  But the combination
$r_1+\omega^2r_2+\omega r_3=3b\sqrt[3]2$
would also be constructible, which might pose a minor problem unless $b=0$.  A similar contradiction using the combination $r_1+\omega r_2+\omega^2r_3$ similarly forces $c=0$, and so the only constructible blnumvers contained in my set are just the rational numbers.
Now try a similar method with your numbers.  All complex fifth roots of unity, like the cube roots of unity described above, are constructible.
