TITLE: What's the explanation for these (infinitely many?) Ramanujan-type identities?
QUESTION [14 upvotes]: Define the function,
$$F(\beta) := \sqrt[3]{\beta+x_1}+\sqrt[3]{\beta+x_2}+\sqrt[3]{\beta+x_3}\tag1$$
where,
$$x_1 =2\cos\big(\tfrac{2\pi }{7}\big),\;x_2 =2\cos\big(\tfrac{4\pi }{7}\big),\;  x_3 =   2\cos\big(\tfrac{8\pi }{7}\big)$$
or the $x_i$ are the roots of the cubic $x^3+x^2-2x-1=0$. We have two nice identities,
$$F(0) = \sqrt[3]{5-3\sqrt[3]7}$$
$$F(1) = \sqrt[3]{-4+3\sqrt[3]7}$$
The first one is by Ramanujan.  (See The Problems Submitted by Ramanujan to the Journal of the Indian Mathematical Society, p. 9, by Bruce Berndt, et al.) 
These two (and other pairs) can be explained by the negative and positive cases of $\pm \sqrt{d}$ in this answer. However, davidoff303 found a plethora of others,
$$F\Big(\frac{74}{43}\Big) = 2\,\sqrt[3]\frac{7^2}{43}$$
$$F\Big(\frac{5105}{11349}\Big) = 3\,\sqrt[3]\frac{7}{11349}$$
$$F\Big(\frac{-2306997866696}{1047656140569}\Big) = -10980\,\sqrt[3]\frac{7^2}{1047656140569}$$
$$F\Big(\frac{9658771264742899051}{5361029308457632889}\Big)= 13\cdot 127\cdot 1381\sqrt[3]{\frac{7}{5361029308457632889}}$$
and so on. Note that, for general rational $\beta$, then $\big(F(\beta)\big)^3$ is a root of a $9$th deg equation.

Q1: Are there infinitely many rational $\beta$ such that $\big(F(\beta)\big)^3$ is also rational, like the ones found by davidoff303? Is there a formula for $\beta$?

$\color{blue}{Update:}$
I formed the $9$th deg equation satisfied by $\big(F(\beta)\big)^3$. Acting on a hunch, it turns out that the $\beta$ used by davidoff303 satisfies the simple relation,
$$\beta^3-\beta^2-2\beta+1 =w^3\tag2$$
This can be transformed into an elliptic curve, and it has infinitely many rational points. And the $LHS$ of $(2)$ in fact is a factor of the discriminant of the $9$th deg eqn.

Q2: Is it true that if rational $\beta$ satisfies $(2)$, then $\big(F(\beta)\big)^3$ is also rational?

REPLY [2 votes]: This is a long comment/addendum to mercio’s answer. Given rational solutions to,
$$\beta^3-\beta^2-2\beta+1 = w^3\tag1$$
As pointed out by mercio, if we define,
$$X = 3\beta^2-2\beta-2+(3\beta-1)w+3w^2 \\Y = 9\beta^3-9\beta^2-11\beta+3w(3\beta^2-2\beta-2)+3w^2(3\beta-1)$$
then this obeys the elegant relation,
$$(Y+2)(Y+9) = X^3$$ 
However, if we require $\beta$ such that
$$Y+2 = u^3\tag2$$
$$Y+9 = v^3\tag3$$
holds separately, then we can define $\big(F(\beta)\big)^3$ as,
$$\big(F(\beta)\big)^3 = \frac{3(t+19)(u+v)+(3uv+7)(27u^4v-63u^3+105uv-196)}{t+19}\tag4$$
where,
$$t=27(u^3+3)(u^3+4)$$
Thus, if $\beta$ satisfies the rational Diophantine conditions $(2),(3)$ in addition to $(1)$, then $\big(F(\beta)\big)^3$ is rational.
Example. Let $\beta = \tfrac{74}{43}$, we get $u=1,\;v=2,\;t=540,$ and using formula $(4)$,
$$\big(F(\tfrac{74}{43})\big)^3 = \frac{2^3\cdot7^2}{43}$$
the same as in the post above. All six examples in the post obey $(1)$, but only the last four obey all $(1), (2),(3)$. Presumably there are infinitely many rational $\beta$ that obey all three conditions, but I do not know how to prove it so.