I was told in another post that Galois theory could help you solve solvable polynomials, and that solvable polynomials had roots that could be expressed as functions of rational numbers, but that it couldn't help me solve a polynomial with rational roots. It cannot help me solve polynomials with roots that couldn't be expressed with radicals. So doesn't that leave me with only polynomials with roots that ARE expressed in radical that Galois theory can help me solve? In other words, does a polynomial's roots HAVE to contain at least one radical for Galois theory to help you solve it?
Polynomials – When Can Galois Theory Help Find Roots of a Polynomial?
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How can one feel comfortable with non-solvable algebraic numbers?
The nice thing about solvable numbers is this idea that they have a formula. You can manipulate the formula as if it were actually a number using some algebra formalism that you probably have felt comfortable with for a while. For instance $\sqrt{3+\sqrt{6}}+2$ is an algebraic number. What do you get if you add it to $7$? Well $\left(\sqrt{3+\sqrt{6}}+2\right)+7=\sqrt{3+\sqrt{6}}+9$ seems like a decent answer. As a side note: there actually some reasonably hard algorithmic questions along these lines, but I'll assume they don't worry you. :-)
We'd like to be able to manipulate other algebraic numbers with similar comfort. The first method I was taught is pretty reasonable:
Kronecker's construction: If $x$ really is an algebraic number, then it is the root of some irreducible polynomial $x^n - a_{n-1} x^{n-1} - \ldots - a_1 x - a_0$. But how do we manipulate $x$? It's almost silly: we treat it just like $x$, and add and multiply as usual, except that $x \cdot x^{n-1}$ needs to be replaced by $a_{n-1} x^{n-1} + \ldots + a_1 x + a_0$, and division is handled by replacing $1/x$ with $( x^{n-1} - a_{n-1} x^{n-2} - \ldots - a_2 x - a_1)/a_0$. This is very similar to "integers mod n" where you replace big numbers by their remainder mod n,. In fact this is just replacing a polynomial in $x$ with its remainder mod $x^n - a_{n-1} x^{n-1} - \ldots - a_1 x - a_0$.
I found it somewhat satisfying, but in many ways it is very mysterious. We use the same symbol for many different algebraic numbers; each time we have to keep track of the $f(x)$ floating in the background. Also it raises deep questions about how to tell two algebraic numbers apart. Luckily more or less all of these questions have clean algorithmic answers, and they are described in Cohen's textbooks CCANT (A Course in Computational Algebraic Number Theory, Henri Cohen, 1993).
Companion matrices: But years later, it still bugged me. Then I studied splitting fields of group representations. The crazy thing about these fields is that they are subrings of matrix rings. So “numbers” were actually matrices. You've probably seen some tricks like this $$\mathbb{C} = \left\{ \begin{bmatrix} a & b \\ -b & a \end{bmatrix} : a,b \in \mathbb{R} \right\}$$ where we can make a bigger field out of matrices over a smaller field. It turns out that is always true: If $K \leq F$ are fields, then $F$ is a $K$-vector space, and the function $f:F \to M_n(K) : x \mapsto ( y \mapsto xy )$ is an injective homomorphism of fields, so that $f(F)$ is a field isomorphic to $F$ but whose “numbers” are just $n \times n$ matrices over $K$, where $n$ is the dimension of $F$ as a $K$-vector space (and yes $n$ could be infinite if you want, but it's not).
That might seem a little complicated, but $f$ just says "what does multiplying look like?" For instance if $\mathbb{C} = \mathbb{R} \oplus \mathbb{R} i$ then multiplying $a+bi$ sends $1$ to $a+bi$ and $i$ to $-b+ai$. The first row is $[a,b]$ and the second row $[-b,a]$. Too easy.
Ok, fine, but that assumes you already know how to multiply, and perhaps you are not yet comfortable enough to multiply non-solvable algebraic numbers! Again we use the polynomial $x^n - a_{n-1} x^{n-1} - \ldots - a_1 x - a_0$, but this time as a matrix. We use the same rule, viewing $F=K \oplus Kx \oplus Kx^2 \oplus \ldots \oplus Kx^{n-1}$ and ask what $x$ does to each basis element: well $x^i$ is typically sent to $x^{i+1}$. It's only the last one that things get funny:
$$f(x) = \begin{bmatrix} 0 & 1 & 0 & 0 & \ldots & 0 & 0 \\ 0 & 0 & 1 & 0 & \ldots & 0 & 0 \\ 0 & 0 & 0 & 1 & \ldots & 0 & 0 \\ & & & & \ddots & & \\ 0 & 0 & 0 & 0 & \ldots & 1 & 0 \\ 0 & 0 & 0 & 0 & \ldots & 0 & 1 \\ a_0 & a_1 & a_2 & a_3 & \ldots & a_{n-2} & a_{n-1} \end{bmatrix}$$
So this fancy “number” $x$ just becomes a matrix, most of whose entries are $0$. For instance $x^2 - (-1)$ gives the matrix $i = \left[\begin{smallmatrix} 0 & 1 \\ -1 & 0 \end{smallmatrix}\right]$.
This nice part here is that different algebraic numbers can actually have different matrix representations. The dark part is making sure that if you have two unrelated algebraic numbers that they actually multiply up like a field. You see $M_n(K)$ has many subfields, but is not itself a field, so you have to choose matrices that both lie within a subfield. Now for splitting fields and centralizer fields and all sorts of handy dandy fancy fields, you absolutely can make sure everything you care about comes from the field. Starting from just a bunch of polynomials though, you need to be careful and find a single polynomial that works for both. This is called the primitive element theorem.
This also lets you see the difference between eigenvalues in the field $K$ and eigenvalues (“numbers”) in the field $F$: the former are actually numbers, or multiples of the identity matrix, while the latter are full-fledged matrices that happen to lie in a subfield. If you ever studied the “real form” of the eigenvalue decomposition with $2\times 2$ blocks, those $2 \times 2$ blocks are exactly the $\begin{bmatrix}a&b\\-b&a\end{bmatrix}$ complex numbers.
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 $d_i := \deg g_i$, $1 \leq i \leq r$.
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 $f_6$.
Best Answer
Let's look at an example. Does Galois Theory help you solve $x^2-3x+1=0$? In a sense, it does; you could use Galois Theory to analyze that equation and eventually to write down its solutions. But you'd be insane to do that: it's ever so much simpler to just write down the answers given by the quadratic formula. Moreover, there is no way that someone who doesn't already know the quadratic formula is going to be able to understand the Galois Theory approach to this equation.
Now for a polynomial whose roots are all rational, it's even worse. The Galois group is trivial, and it basically tells you that the way to find the roots is to use the rational root theorem.
So I could say, yes, Galois Theory helps you solve the equation; it helps you by telling you to use the rational root theorem. You might reply, Galois Theory doesn't help you solve the equation; all it does it tell you to use the rational root theorem. We would agree on the mathematics, and disagree on the interpretation. I don't think we can take it much further than that.