, 1997

Let $\alpha = \cos \frac{\pi}{7}$. As $\cos \frac{4\pi}{7} + \cos \frac{3\pi}{7} = 0$, we have
$$2(2\alpha^2 - 1)^2 - 1 + (4\alpha^3 - 3\alpha) = 0.$$
This is the same as $(\alpha + 1)(8\alpha^3 - 4\alpha^2 - 4\alpha + 1) = 0$. Clearly, $\alpha \ne -1$. Therefore, $\alpha$ is a root of
$$P(x) = 8x^3 - 4x^2 - 4x + 1.$$
Suppose on the contrary that $\alpha = p + \sqrt{q} + \sqrt[3]{r}$. Then $\alpha$ is also a root of
$$Q(x) = (x - p - \sqrt{q})^3 - r = x^3 - 3(p + \sqrt{q})x^2 + 3(p + \sqrt{q})^2 x - (p + \sqrt{q})^3 - r.$$
Consider the field $F = \{a + b\sqrt{q} : a, b \in \mathbb{Q}\}$ (which is just $\mathbb{Q}$ if $q = 0$). We claim that $P(x)$ has a root in $F$.
Firstly, note that $P(x) \ne 8Q(x)$. Indeed, if $P(x) = 8Q(x)$, we must have $q = 0$ in order that all coefficients of $Q(x)$ are rational. Then by comparing the coefficients of $x^3$ and $x^2$, we need $p = \frac{1}{6}$. But then the coefficients of $x$ do not match. Therefore, $R(x) = \gcd(P(x), Q(x))$ (over $F[x]$) has degree 1 or 2.

* If $\deg R = 1$, then since $\alpha$ is a root of $P(x)$ and $Q(x)$, it is a root of $R(x)$, and hence belongs to $F$.
* If $\deg R = 2$, we can write $P(x) = R(x)S(x)$ where $\deg S = 1$. Then $S(x)$ has a root in $F$, which is also a root of $P(x)$.
Now, let $\beta = a + b\sqrt{q}$ be a root of $P(x)$ in $F$. Then $\beta$ is a root of $(x-a)^2 = b^2q$, and so the minimal polynomial of $\beta$ over $\mathbb{Q}$ has degree at most 2. As this minimal polynomial divides $P(x)$, we find that $P(x)$ is reducible. As $\deg P = 3$, it must consist of a linear factor. Thus, $P(x)$ has a rational root. However, it is routine to check that none of
$$\pm 1, \pm \frac{1}{2}, \pm \frac{1}{4}, \pm \frac{1}{8}$$
is a root of $P(x)$. By the rational root theorem, $P(x)$ does not have any rational root. This is a contradiction. Therefore, $\alpha$ cannot be expressed in the form $p + \sqrt{q} + \sqrt[3]{r}$.