Solving a quartic polynomial without using the cube root operator I have been learning Galois theory and was curious about this question. Given a quartic polynomial, when can we write down the roots of it without using any cube roots? Does some condition on the coefficients of the polynomial or the Galois group of the polynomial imply this condition? 
Thanks!
 A: Let $f \in K[X]$ be your quartic polynomial, with splitting field $L$. Your condition forces $L$ to arise from $K$ by successive quadratic extensions and therefore $\mathrm{Gal}(L/K)$ is a $2$-group. Conversely, as long as $\mathrm{char}(K) \ne 2,$ such a Galois group implies that $f$ can be solved by quadratic radicals.
Assuming $f$ is irreducible, this limits the possible Galois groups to $D_8$, $V_4 = \mathbb{Z}/2 \times \mathbb{Z}/2$ or $\mathbb{Z}/4$. Writing $$f(X) = X^4 + aX^3 + bX^2 + cX + d,$$ you are in one of those three cases if and only if the resolvent cubic $$R_f(X) = X^3 - bX^2 + (ac - 4d) X - (a^2 d - 4bd + c^2)$$ is reducible. So in this sense you can read it off the coefficients.
A: here are two good ones
$$ x^4 + x^3 + 2 x^2 - 4 x + 3  $$
$$ x^4 + x^3 -6 x^2 -  x + 1  $$
For the first one, if $\omega$ is any 13th root of unity (but not one itself), then $\omega + \omega^3 + \omega^9 $ is a root.   For the second one, if $\omega$ is any 17th root of unity (but not one itself), then $\omega + \omega^4 + \omega^{16}  + \omega^{64} =\omega + \omega^4 + \omega^{-1}  + \omega^{-4} $ is a root.
