Is $f(x) = x^{10}-x^5+1$ solvable by radicals?

Question

Is $f(x) = x^{10}-x^5+1$ solvable by radicals?

So far I've showed that $f$ is irreducible because if we let $y=x^5$ then $f(y)=y^2-y+1$ which is irreducible because it has a negative discriminant. I also know that $f$ has no real roots so I've concluded that $Gal(L_f/\mathbb{Q}) \subset S_{10}$ and that there is a 10-cycle and a transposition in $Gal(L_f/\mathbb{Q})$. However I have no clue as to what $Gal(L_f/\mathbb{Q})$ could be.

Answer 1

If you multiply by $x^5+1$, you get $x^{15}+1 =0$. So the roots are all the 15th roots of $-1$. Throw out the five 5th-roots, and you have your solutions. So the answer is "yes".

Answer 2

Yes, of course.

Let $x+\frac{1}{x}=a$.

Hence, $$x^{10}-x^5+1=x^{10}+x^7-x^7+x^6-x^5-x^6+1=$$ $$=(x^2-x+1)(x^7(x+1)-x^5-(x^3-1)(x+1))=$$ $$=(x^2-x+1)(x^8+x^7-x^5-x^4-x^3+x+1)=$$ $$=(x^2-x+1)x^4\left(x^4+\frac{1}{x^4}+x^3+\frac{1}{x^3}-x-\frac{1}{x}-1\right)=$$ $$=(x^2-x+1)x^4(a^4-4a^2+2+a^3-3a-a-1)=$$ $$=(x^2-x+1)x^4(a^4+a^3-4a^2-4a+1)=$$ $$=(x^2-x+1)x^4\left(a^2+\frac{1-\sqrt5}{2}a+\frac{3+\sqrt5}{2}\right)\left(a^2+\frac{1+\sqrt5}{2}a+\frac{-3+\sqrt5}{2}\right)...$$