Proof of impossibility
I write this section for the following reasons:
- A proof of impossibility is much more interesting than solving it in terms of $\cos$.
- You tagged galois-theory, which is also largely motivated by the concept of proving something is impossible.
- Casus irreducibilis is mentioned in the comments.
I will attempt to formalize the problem before proving its impossibility:
Is there an intermediate field $F$ between $\Bbb Q$ and $\Bbb R$ so that $f(X) = X^3 - X^2 - 2X + 1$ splits in $F$ and $F$ is radical over $\Bbb Q$?
Explanation
I will explain how the idea of "solvable by radicals" is captured by the idea of "radical extension".
For example, if we want to find the number $\sqrt[3]{11} \sqrt[5]{\dfrac{7+\sqrt3}2} + \sqrt[4]{1+\sqrt[3]4}$ (example given by Ian Stewart in his third edition of Galois Theory), we might want to break this number down and extend our field $\Bbb Q$ adjoining radicals one by one, i.e. come up with:
$$\Bbb Q \subset \Bbb Q\left(\sqrt[3]{11}, \sqrt3, \sqrt[5]{\dfrac{7+\sqrt3}2},\sqrt[3]4,\sqrt[4]{1+\sqrt[3]4}\right) = \Bbb Q(a,b,c,d,e)$$
What makes this extension radical is the fact that each adjoined element, raised to some power, can be found in the previous field:
- From $\Bbb Q$ to $\Bbb Q(a)$, we find that $a^3 = 11 \in \Bbb Q$.
- From $\Bbb Q(a)$ to $\Bbb Q(a,b)$, we find that $b^2 = 3 \in \Bbb Q(a)$.
- From $\Bbb Q(a,b)$ to $\Bbb Q(a,b,c)$, we find that $c^5 = \dfrac{7+b}2 \in \Bbb Q(a,b)$.
- From $\Bbb Q(a,b,c)$ to $\Bbb Q(a,b,c,d)$, we find that $d^3 = 4 \in \Bbb Q(a,b,c)$.
- From $\Bbb Q(a,b,c,d)$ to $\Bbb Q(a,b,c,d,e)$, we find that $e^4 = 1+d \in \Bbb Q(a,b,c,d)$.
Note that a radical extension may not be a normal extension, i.e. an extension where every irreducible polynomial over the base field that has at least one zero in the extension splits in the extension. For example, $\Bbb Q(\sqrt[4]2)$ is not a normal extension of $\Bbb Q$, because $X^4-2$ is a polynomial in $\Bbb Q[X]$ that is irreducible, but only two of its roots are in $\Bbb Q(\sqrt[4]2)$ (the other two being $i\sqrt[4]2$ and $-i\sqrt[4]2$).
Proof
A large portion of this proof is inspired by Theodore Shifrin's proof in his book (see section Bibliography).
If such an intermediate field $F$ exists, according to the definition of a radical extension, write it as a tower of extensions:
$$\Bbb Q \subset \Bbb Q(\sqrt[p_1]{b_1}) \subset \Bbb Q(\sqrt[p_1]{b_1}, \sqrt[p_2]{b_2}) \subset \cdots \subset K \subset K(\sqrt[p]a) = F \subset \Bbb R$$
where $b_{i+1} \in \Bbb Q(\sqrt[p_1]{b_1}, \sqrt[p_2]{b_2}, \cdots, \sqrt[p_i]{b_i})$ for each $i$.
WLOG, each $p_i$ is a prime, or else we factorize the composite number and include more intermediate fields to the tower such that each $p_i$ is a prime.
WLOG, let $a_i$ be not a $p_i$th power, or else we can rightfully remove that field from the tower, as it would just be the same as the previous field.
WLOG, let the polynomial $f(X)$ be irreducible in $K$ but has a root in $K(\sqrt[p]a)$ (or else, find such a point in the tower where the previous field has no roots for the polynomial but the next one does, and cut the tower there).
From the other answers, the polynomial has roots in terms of cosine, and one can see that one can write the other roots in terms of one root, using the multiple angle formulas for cosine. Therefore, a field containing one root must contain all of the roots.
Therefore, the polynomial actually splits in $K(\sqrt[p]a)$.
Let the splitting field of the polynomial over $K$ be $L$. We have:
$$K \subset L \subset K(\sqrt[p]a)$$
Then, $[L:K] = 3$ as the polynomial has degree $3$. From Lemma 4.7 of Generalizations of the Casus Irreducibilis (link to download), we know that $[K(\sqrt[p]a):K] = p$. Therefore, $p$ is divisible by $3$. But since $p$ is prime, we have $p=3$, whence $L = K(\sqrt[3]a)$.
Since $a$ is not a $3$th power in $K$, $X^3-a$ is irreducible over $K$ (why?). We will prove that $\omega\sqrt[3]a \in L$. (This portion of the proof is highly motivated by the proof that Ian Stewart gave in his Galois Theory.)
Note that $[K(\omega\sqrt[3]a):K] = [K(\sqrt[3]a):K] = 3$, for their minimal polynomial over $K$ is the same ($X^3 - a$). Any isomorphism from $L(\sqrt[3]a)$ to $L(\omega\sqrt[3]a)$ defines an isomorphism from $K(\sqrt[3]a)$ to $K(\omega\sqrt[3]a)$, so $[L(\sqrt[3]a):K(\sqrt[3]a)] = [L(\omega\sqrt[3]a):K(\omega\sqrt[3]a)]$. Therefore:
$$\begin{array}{rcll}
[L(\omega\sqrt[3]a):L]
&=& \dfrac{[L(\omega\sqrt[3]a):K]}{[L:K]} & \text{tower law} \\
&=& \dfrac{[L(\omega\sqrt[3]a):K(\omega\sqrt[3]a)][K(\omega\sqrt[3]a):K]}{[L:K]} & \text{tower law} \\
&=& \dfrac{[L(\sqrt[3]a):K(\sqrt[3]a)][K(\sqrt[3]a):K]}{[L:K]} \\
&=& \dfrac{1 \times 3}{3} \\
&=& 1
\end{array}$$
Therefore, $L(\omega\sqrt[3]a) = L$, so $\omega\sqrt[3]a \in L$, which contradicts $L = K(\sqrt[3]a) \subset \Bbb R$.
$\blacksquare$
Bibliography
The following two links are kept here although it is no longer used above.
Roots expressed in terms of other roots
From the answers above, we know that the roots can be expressed as $a=2\cos\left(\dfrac\pi7\right)$, $b=2\cos\left(\dfrac{3\pi}7\right)$, and $c=2\cos\left(\dfrac{5\pi}7\right)$.
In this section, I will express the roots in terms of the other roots, leaving us with $9$ expressions (among which $3$ are trivial).
Unless otherwise indicated, the formulas are obtained via multiple angle formulas, and the fact that all roots satisfy $x^3 - x^2 - 2x + 1 = 0$.
Express in terms of $a$:
We note that $c=-2\cos\left(\dfrac{2\pi}7\right)$.
- $a=a$
- $b=2\left[4\left(\dfrac a2\right)^3-3\left(\dfrac a2\right)\right]=a^2-a-1$
- $c=-2\left[2\left(\dfrac a2\right)^2-1\right]=-a^2+2$
Express in terms of $b$:
We note that $a=-2\cos\left(\dfrac{6\pi}7\right)$ and $c=2\cos\left(\dfrac{9\pi}7\right)$.
- $a=-2\left[2\left(\dfrac b2\right)^2-1\right]=-b^2+2$
- $b=b$
- $c=2\left[4\left(\dfrac b2\right)^3-3\left(\dfrac b2\right)\right]=b^2-b-1$
Express in terms of $c$:
We note that $a=-2\cos\left(\dfrac{6\pi}7\right)$, $b=-2\cos\left(\dfrac{4\pi}7\right)$, and $c=-2\cos\left(\dfrac{2\pi}7\right)$.
- $a=-2\left[4\left(-\dfrac c2\right)^3-3\left(-\dfrac c2\right)\right]=c^2-c-1$
- $b=-2\left[2\left(-\dfrac c2\right)^2-1\right]=-c^2+2$
- $c=c$
Solution (allowing complex numbers)
For the sake of completeness, a solution including complex numbers is included.
Let $p(x) = x^3 - x^2 - 2x + 1$.
The discriminant of $p$ is $49 = 7^2$, so the Galois group of the splitting field of $p$ over $\Bbb Q$ is $\Bbb A_3$.
Let the roots of $p$ be $a$, $b$, and $c$.
From Vieta:
$$\begin{cases}
a+b+c &=& 1 \\
ab+bc+ac &=& -2 \\
abc &=& -1
\end{cases}$$
We find $a^2b+b^2c+c^2a+ab^2+bc^2+ca^2$:
$$\begin{array}{rcl}
(a+b+c) (ab+bc+ac) &=& -2 \\
a^2b+b^2c+c^2a+ab^2+bc^2+ca^2+3abc &=& -2 \\
a^2b+b^2c+c^2a+ab^2+bc^2+ca^2 &=& 1 \\
\end{array}$$
We find $a^3+b^3+c^3$:
$$\begin{array}{rcl}
(a+b+c)^3 &=& 1 \\
a^3+b^3+c^3 + 3(a^2b+b^2c+c^2a+ab^2+bc^2+ca^2) + 6abc &=& 1 \\
a^3+b^3+c^3 + 3 - 6 &=& 1 \\
a^3+b^3+c^3 &=& 4 \\
\end{array}$$
We find $a^3b^3+b^3c^3+c^3a^3$:
$$\begin{array}{rcl}
(ab+bc+ac)^3 &=& a^3b^3+b^3c^3+c^3a^3 + 3(ab+bc+ac)(a^2bc+ab^2c+abc^2) - 3(abc)^2 \\
-8 &=& a^3b^3+b^3c^3+c^3a^3 + 3(ab+bc+ac)(abc)(a+b+c) - 3 \\
-8 &=& a^3b^3+b^3c^3+c^3a^3 + 3(-2)(-1)(1) - 3 \\
a^3b^3+b^3c^3+c^3a^3 &=& -11 \\
\end{array}$$
We find $(a^2b+b^2c+c^2a)(ab^2+bc^2+ca^2)$:
$$\begin{array}{cl}
& (a^2b+b^2c+c^2a)(ab^2+bc^2+ca^2) \\
=& a^3b^3+b^3c^3+c^3a^3+a^4bc+b^4ca+c^4ab+3a^2b^2c^2 \\
=& a^3b^3+b^3c^3+c^3a^3+abc(a^3+b^3+c^3)+3(abc)^2 \\
=& -11+(-1)(4)+3(-1)^2 \\
=& -12
\end{array}$$
Let $h = a^2b+b^2c+c^2a$ and $k = a^2b+bc^2+ca^2$. From the above, $h+k=1$ and $hk=-12$.
Let $p = a+b\omega+c\omega^2$. The Galois group permutes $a$, $b$, and $c$, changing the expression to $p\omega = c+a\omega+b\omega^2$ and $p\omega^2 = b+c\omega+a\omega^2$.
Therefore, $p^3$ is fixed:
$$\begin{array}{rcl}
p^3
&=& (a+b\omega+c\omega^2)^3 \\
&=& a^3 + b^3 + c^3 + 3(a^2b\omega + ab^2\omega^2 + a^2c\omega^2 + ac^2\omega + b^2c\omega + bc^2\omega^2) + 6abc \\
&=& a^3 + b^3 + c^3 + 3\omega h + 3\omega^2 k + 6abc \\
&=& 4 + 3\omega h + 3\omega^2 k - 6 \\
&=& -2 + 3\omega h + 3\omega^2 k \\
\end{array}$$
Let $q = a+b\omega^2+c\omega$. By the same argument, $q^3$ is fixed:
$$\begin{array}{rcl}
q^3
&=& (a+b\omega^2+c\omega)^3 \\
&=& a^3 + b^3 + c^3 + 3(a^2b\omega^2 + ab^2\omega + a^2c\omega + ac^2\omega^2 + b^2c\omega^2 + bc^2\omega) + 6abc \\
&=& a^3 + b^3 + c^3 + 3\omega^2 h + 3\omega k + 6abc \\
&=& 4 + 3\omega^2 h + 3\omega k - 6 \\
&=& -2 + 3\omega^2 h + 3\omega k \\
\end{array}$$
Then:
$$\begin{array}{rcl}
p^3 + q^3
&=& -4 + 3(\omega + \omega^2) (h+k) \\
&=& -4 + 3(-1)(1) \\
&=& -7 \\
\end{array}$$
Also:
$$\begin{array}{rcl}
p^3 q^3
&=& 4 + 9h^2 + 9k^2 - 6(\omega+\omega^2)h - 6(\omega+\omega^2)k + 9(\omega^2+\omega)hk \\
&=& 4 + 9[(h+k)^2-2hk] - 6(-1)(h+k) + 9(-1)hk \\
&=& 4 + 9[1^2-2(-12)] - 6(-1)(1) + 9(-1)(-12) \\
&=& 4 + 225 + 6 + 108 \\
&=& 343
\end{array}$$
Therefore, $p^3-q^3 = \sqrt{(p^3+q^3)^2-4p^3q^3} = \sqrt{-1323} = 21i\sqrt{3}$.
We obtain $p^3 = \dfrac12[(p^3+q^3)+(p^3-q^3)] = \dfrac{-7+21i\sqrt{3}}2$ and $q^3 = \dfrac12[(p^3+q^3)-(p^3-q^3)] = \dfrac{-7-21i\sqrt{3}}2$.
Therefore:
- $(1): \quad a+b+c=1$ (by Vieta)
- $(2): \quad a+b\omega+c\omega^2=\sqrt[3]{\dfrac{-7+21i\sqrt{3}}2}$
- $(3): \quad a+b\omega^2+c\omega=\sqrt[3]{\dfrac{-7-21i\sqrt{3}}2}$
From $\dfrac13[(1)+(2)+(3)]$:
$$a = \frac13 + \frac13\sqrt[3]{\dfrac{-7+21i\sqrt{3}}2} + \frac13\sqrt[3]{\dfrac{-7-21i\sqrt{3}}2}$$
From $\dfrac13[(1)+(2)\omega^2+(3)\omega]$:
$$b = \frac13 + \frac13\frac{-1-i\sqrt3}2\sqrt[3]{\dfrac{-7+21i\sqrt{3}}2} + \frac13\frac{-1+i\sqrt3}2\sqrt[3]{\dfrac{-7-21i\sqrt{3}}2}$$
From $\dfrac13[(1)+(2)\omega+(3)\omega^2]$:
$$c = \frac13 + \frac13\frac{-1+i\sqrt3}2\sqrt[3]{\dfrac{-7+21i\sqrt{3}}2} + \frac13\frac{-1-i\sqrt3}2\sqrt[3]{\dfrac{-7-21i\sqrt{3}}2}$$
P.S.: $\sqrt[3]{re^{i\theta}} = r^{1/3} e^{i\theta/3}$, where $\theta \in (-\pi,\pi]$.
