Let $f(x) = x^{12}+2x^6-2x^3+2$ and let $K$ be it's splitting field. Is the Galois group of $K/Q$ solvable?
Since we're really only interested in the polynomial $x^4+2x^2-2x+2$, and we know that there exists a quartic formula. It is enough to conclude that the polynomial is solvable by radicals and therefore the Galois group is solvable. Is this solution correct?
The question also asks if this polynomial is irreducible. This is easy to show via Eisenstein's criterion. But is the irreducibility of the polynomial important in any way for showing it is solvable?
The Galois group of $f(x)$ has order $3888=2^4\cdot 3^5$. For the computation one can use Magma online here. This group is, by Burnside's $p^rq^s$-Theorem solvable, since the order has only two different prime divisors. The polynomial $x^4+2x^2-2x+2$ has a solvable Galois group, see K. Conrad's article Galois group of cubics and quartics, namely one of the following solvable groups: $S_4,A_4,D_4,C_4,C_2\times C_2$. However, the Galois group of $x^{12}+2x^6-2x^3+2$ is quite different from these groups.