Can this equation be solved by radicals? $x^6+x^5+2x^4-2x^3-x^2+1=0$

Question

Can this equation be solved by radicals? If it can be solved, what is the method? At least, I will resolve it myself if you specify reference or source.

$$x^6+x^5+2x^4-2x^3-x^2+1=0$$

Answer 1

Yes because$$x^6+x^5+2x^4-2x^3-x^2+1=0\iff (x^3-1)^2+x^2(x^3-1)+2x^4=0$$ It follows solving as a quadratic in $x^3-1$ $$2(x^3-1)=-x^2\pm\sqrt{-7}x\iff2x^3+x^2\mp\sqrt{-7}x^2-2=0$$ so you can apply Cardano's formula.

Answer 2

Let $a=\frac{1+\sqrt{-7}}{2}$, $E=Q(a)$, then the given polynomial factors as $f=(x^3+ax^2-1)(x^3+\bar{a}x^2-1)$. It is easy to see that the cubic factors are irreducible over $E$ and that their discriminants are not squares in $E$. Hence the Galois group of $f$ over $E$ is $H=S_3\times S_3$ and that the Galois group $G$ of the original polynomial is the semidirect product extension of $H$ by $Z_2$ (complex conjugation). Thus the full Galois group $G$ is of order 72 and solvable. To obtain the roots we can use the the standard formulas for solving a cubic.