I don't know why I'm having trouble with this, but I can't quite see whether the cyclotomic polynomials are considered solvable. Obvioulsy we can write the solution of the nth cyclotomic polynomial as the nth root of unity, which seems to be a perfectly good algebraic solution; but aren't there usually "better" solutions? I've done the fifth and seventh, and I think Gauss is credited for solving the eleventh if I'm not mistaken; the seventeenth falls out of the construction of the regular 17-gon…but I'm still not clear: are all of them, in theory, solvable "explicitly", which is to say beyond the level of just saying that the nth root of unity is a solution?
Abstract Algebra – Solvability of Cyclotomic Polynomials
abstract-algebrapolynomials
Best Answer
A polynomial is solvable iff its splitting field is solvable iff its Galois group is solvable. The Galois group of the cyclotomic polynomial $\Phi_n(x)$ is $(\mathbb{Z}/n\mathbb{Z})^{\ast}$, which is abelian, and all abelian groups are solvable.
An interesting question is how to find "nice" radical expressions for roots of unity. One way to do this is to go through the proof that the roots of a solvable polynomial are expressible in radicals by finding a composition series of the Galois group and constructing Kummer extensions.
This is perhaps easiest to describe by example, so take $n = 5$. Then $\Phi_5(x) = x^4 + x^3 + x^2 + x + 1$ has Galois group $(\mathbb{Z}/5\mathbb{Z})^{\ast} \cong C_4$, so it has a composition series with two factors of $C_2$. This implies that $\mathbb{Q}(\zeta_5)$ is a quadratic extension of a quadratic extension. Basic facts about Gauss sums imply that the second quadratic extension is $\mathbb{Q}(\sqrt{5})$, so $\zeta_5$ satisfies a quadratic polynomial with coefficients in $\mathbb{Q}(\sqrt{5})$. This polynomial must be $$x^2 - (\zeta_5 + \zeta_5^{-1}) x + 1.$$
Again, basic facts about Gauss sums imply that $\zeta_5 + \zeta_5^{-1} = \frac{-1 + \sqrt{5}}{2}$, so it follows by the quadratic formula that $$\zeta_5 = \frac{ 1 - \sqrt{5} + i \sqrt{10 + 2 \sqrt{5}} }{4}.$$
In general one may need to take more than just square roots, in which case things get complicated. There will be expressions one can write down generalizing Gauss sums that are guaranteed to land in a specific subfield of $\mathbb{Q}(\zeta_n)$ but I don't know a good way of actually figuring out what those expressions are by hand, and in writing down Kummer extensions one may need to adjoin smaller roots of unity (so I believe the most important case is when $n$ is prime).
A basic remark is that the problem reduces to the case that $n$ is a power of a prime, since for arbitrary $n$ it is possible to write $\zeta_n$ as a product of roots of the form $\zeta_{p^k}$ where $p^k \parallel n$.