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

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