Can the roots to the polynomial $x^6 - x^5 - x^4 - x^3 - x^2 - x - 1$ be found in terms of radicals? I have been trying to derive a closed-form expression for a linear recurrence relation (this was originally motivated by an attempt to reduce the time complexity on a competitive programming question, whose naive solution is a Dynamic Programming approach).
In doing so, the following sextic polynomial appears as the characteristic polynomial of the 6x6 matrix describing the recurrence transform:
$x^6 - x^5 - x^4 - x^3 - x^2 - x - 1$ 
The roots of this polynomial can be easily approximated, and the approximate closed-form expression follows, but I was wondering – could the exact roots could be derived, and expressed in terms of radicals?
I know only certain classes of sextics can be solved in terms of radicals; but as I am not too familiar with Galois Theory in any quantitative sense whatsoever, I haven't been able to determine if the above is such a polynomial.
Indeed, if the polynomial cannot be expressed in radical terms, could an exact solution be represented by alternative means (ideally, means generalizable to other sextic or $n$th degree polynomials)?
Thanks.
 A: Answers to How can I tell if $x^5 - (x^4 + x^3 + x^2 + x^1 + 1)$ is/is not part of the solvable group of polynomials? use a dedicated CAS function (e.g., Maple's $\texttt{galois()}$) to compute that the Galois group $\operatorname{Gal}(f_n)$ of $f_n(x) := x^n - \sum_{i = 0}^{n - 1} x^i$ is the full group $S_n$ for $n \leq 12$. For $n \geq 5$, $S_n$ is not solvable, and so for $5 \leq n \leq 12$ the roots of $f_n$ cannot be expressed in terms of radicals. In particular this applies to the polynomial in this question:
$$\color{#df0000}{\boxed{\textrm{The roots of $x^6 - x^5 - x^4 - x^3 - x^2 - x - 1$ cannot be expressed in terms of radicals.}}}$$
To show this result for $f_6$ without using $\mathtt{galois()}$ or a similar function one can proceed as follows.
Proof. First observe that $f_6$ is irreducible modulo $5$, hence irreducible over $\mathbb Z[x]$. Now, recall that

*

*A theorem of Dedekind [pdf] asserts that if $p$ is a prime not dividing the discriminant of a polynomial $f$ and $f(x)$ factors into distinct irreducible polynomials modulo $p$ as $$f(x) \equiv g_1(x) \cdots g_r(x) \pmod p ,$$ then the Galois group $\operatorname{Gal}(f)$ of $f$ (regarded as a subgroup of the symmetric group $S_n$ of permutations of its roots) contains an element of cycle type $(d_1, \ldots, d_r)$, where $d_i := \deg g_i$, $1 \leq i \leq r$.


*If a transitive subgroup $G \leq S_n$—in particular, the Galois group of an irreducible polynomial of degree $n$—contains both a transposition and an $(n - 1)$-cycle, then $G = S_n$.
So, it suffices to use Dedekind's Theorem to find primes $p$ not dividing the discriminant $\Delta := 205\,937$ of $f_6$ for which $f_6$ factors modulo $p$ into products of irreducible polynomials of appropriate degree and then apply (2) to conclude that $\operatorname{Gal}(f_6) = S_n$. Since $\Delta$ is prime, the restriction on $p$ is just $p \neq \Delta$.
First, $f_6(x) \equiv (x + 2) g(x) \pmod {17}$ for an irreducible quintic polynomial $g$, so $\operatorname{Gal}(f_6)$ contains a $5$-cycle. On the other hand, $f_6(x) \equiv (x + 2) (x + 4) h(x) \pmod 5$ for an irreducible quartic polynomial $h$, so $\operatorname{Gal}(f_6)$ contains a product a $4$-cycle $\sigma$, and in particular its square $\sigma^2 \in \operatorname{Gal}(f_6)$ is a transposition. So, by (2) $\operatorname{Gal}(f_6) = S_6$. $\blacksquare$
I'm not aware of any way to express roots of a general sextic polynomial beyond simply naming them, but we can apply a clever transformation (courtesy of achille hui in the comments) to write the roots of this particular sextic in terms of a particular infinite series. Computing $(x - 1) f_6(x)$, substituting $x = 2^{-1 / 6} t^{-1}$ and multiplying through by $t^7$ gives the septic polynomial $t^7 - t + 2^{-7 / 6}$. Now, referring to a derivation of Glasser gives that we can write the roots $t_k$, $k = 1, \ldots, 6$, of the polynomial in $t$ (other than $t_0 = 2^{-1 / 6}$, which corresponds to the $x_0 = 1$) as
$$t_k = e^{- \pi k i / 3} - \frac{1}{2^{13 / 6} \cdot 3} \sum_{\ell = 0}^\infty \frac{e^{\pi k \ell i / 3}}{2^{7 \ell / 6} (\ell + 1)!} \frac{\Gamma\left(\frac{7 \ell}{6} + 1\right)}{\Gamma\left(\frac{\ell}{6} + 1\right)} .$$
Substituting back gives that the roots of $f_6(x)$ are $x_k = 2^{-1 / 6} t_k^{-1}$. The roots $t_k$ (and hence $x_k$) can also be written in terms of no more than six values of hypergeometric functions, but an explicit expression is too tedious to write out here; for more see the link in this paragraph.
The unique positive root, $1.98358\!\ldots$, of $f_6$ is sometimes called the Hexanacci constant. See also the MathWorld article on Hexanacci numbers, a generalization of the Fibonacci numbers, which can be expressed in terms of the roots of $f_6$.
