What's the explanation for these (infinitely many?) Ramanujan-type identities?

Since there are $3$ ways to choose each of the cube roots, in general, $F(\beta)$ is a root of a degree $27$ polynomial whose coefficients are expressions in $\beta,(x_1+x_2+x_3),(x_1x_2,x_2x_3,x_3x_1),(x_1x_2x_3)$.

Since multiplying each root by $\zeta_3$ doesn't change the value of $F(\beta)^3$, this polynomial is in fact a polynomial of degree $9$ in $F(\beta)^3$.

Assume now that we have $(\beta+x_1)(\beta+x_2)(\beta+x_3) = w^3$.

Then the degree $9$ polynomial factors after extending our coefficients field with $w$, into a cubic and a sextic : gather in the cubic the three choices of roots giving $(\beta+x_1)^{1/3}(\beta+x_2)^{1/3}(\beta+x_3)^{1/3} = w$

Then, $F(\beta)^3 = (3\beta+x_1+x_2+x_3) + 6w + 3\sum_{i \neq j} (\beta+x_i)^\frac 23(\beta+x_j)^\frac 13 \\ = (3\beta+x_1+x_2+x_3) + 6w + 3w\sum_{i \neq j} (\beta+x_i)^\frac 13(\beta+x_j)^{-\frac 13}$

And so $(F(\beta)^3$ is rational if and only if the sum is. At this point, a choice of $F(\beta)$ among the $3$ possibilities is equivalent to a choice of only one term if the sum. Or equivalently, you go from one value to the others by multiplying $(\beta+x_2)^\frac 13$ by $\zeta_3$ and $(\beta+x_3)^\frac 13$ by $\zeta_3^2$

Split the sum into two parts $z_1 + z_2$ (one for each coset of $A_3$ in $S_3$). When applying this automorphism, one is multiplied by $\zeta_3$ and the other is multiplied by $\zeta_3^2$, so their product $z_1z_2$ has to be invariant, and indeed, $w^2z_1z_2 = 3\beta^2-2\beta-2+(3\beta-1)w+3w^2$.

If the original Galois group between $x_1,x_2,x_3$ had been $S_3$, things would have been more complicated, but here, since the Galois group is cyclic, $x_1/x_2+ x_2/x_3 + x_3/x_1$ is rational, and this implies that $z_1^3$ and $z_2^3$ are rational :

$\{(wz_i)^3\} = 9\beta^3-9\beta^2-11\beta+3w(3\beta^2-2\beta-2)+3w^2(3\beta-1) + \{2;9\}$

If one of them is $0$, then the other is $\pm 7$, and this corresponds to the $\beta = 0,1$ cases.

If not, then $F(\beta)^3$ is rational if and only if one of those two rational numbers is a cube (since their product is a nonzero cube, this is equivalent to both being a cube).

The map $(\beta,w) \mapsto \\ (X = 3\beta^2-2\beta-2+(3\beta-1)w+3w^2 ; \\ Y = 9\beta^3-9\beta^2-11\beta+3w(3\beta^2-2\beta-2)+3w^2(3\beta-1))$

is a rational isomorphism between the elliptic curves $E_1 : w^3 = (\beta+x_1)(\beta+x_2)(\beta+x_3)$ and $E_2 : (Y+2)(Y+9) = X^3$ (it can't be ramified, and the only preimage of the point at infinity from $E_2$ is the point at infinity with $\beta/w = 1$)

Since $Y+2$ has a triple pole at the point at infinity of $E_2$ and a triple zero at $(X=0,Y=-2)$, $Y+2$ is not a cube (there is no degree $1$ map from an elliptic curve into $\Bbb P^1$), but the extension obtained by adding a cube root of $Y+2$ is a $3$ fold unramified covering $f : E_3 \to E_2 \cong E_1$

Then the rational points on $E_1$ whose $F(\beta)^3$ is rational are (aside from the two exceptions with $X=0$) the images of the rational points on $E_3$.

Choose a rational point on $E_3$ (such as any example where $F(\beta)$ is a cube, or the rational point above the point at infinity of $E_2$) so we can have a group law on $E_3$ and $E_2$.

If $\hat f : E_2 \to E_3$ is the dual map to $f$ then $f \circ \hat f = [3]$, so $f(E_3(\Bbb Q))$ is a subgroup of $E_2(\Bbb Q)$ that has to contain $[3](E_2(\Bbb Q))$

If you name $u = (Y+2)^\frac 13,v = (Y+9)^\frac 13$, then $E_3$ is given by $v^3=u^3+7$ and $f : E_3 \to E_2$ is given by $(X = uv, Y = u^3-2)$. After composing this with the isomorphism $E_2 \to E_1$ (that I didn't compute), you get a nice rational algebraic parametrization by $E_3$ of the points you want, $(u,v) \mapsto (\beta,w,F(\beta)^3)$

From here on, things might be different with other choices of $(x_1,x_2,x_3)$ or other base fields.

According to http://www.lmfdb.org/EllipticCurve/Q/441/d/2

$E_2(\Bbb Q)$ 's torsion subgroup has order $3$ and is generated by $Q_1 = (X=0, Y=-2) = (\beta = 1, w = -1)$. Since $Y+9$ is not a cube, it doesn't have a rational preimage in $E_3$ and $F(1)^3$ is not rational.

The other point of order $3$ is $2Q_1 = Q_2 = (X=0, Y=-9) = (\beta = 0, w = 1)$ and is in the same case.

$E_2(\Bbb Q)$ has rank $1$ and its free part is generated by $R = (X=2,Y= -1) = (\beta = \frac {74}{43}, w = - \frac {29}{43})$. Since $Y+2$ and $Y+9$ are nonzero cubes, it has a rational preimage in $E_3$, and $F(\frac {74}{43})^3$ is rational.

This is enough to determine that the subgroup of "good" points is generated by $(X=2,Y= -1)$. For example $2R = (X = -\frac{20}9, Y = -\frac{118}{27}) = (\beta = \frac {5105}{11349}, w = - \frac {2521}{11349})$, and so on.

Since $E_3$ also has a group law, there are addition formulas that combine triplets of rationals $(\beta,w,F(\beta)^3)$, so if you compute those formulas, you get an easy way to generate the points you want.

As you noticed, $\tau : (\beta,w) \mapsto (-\frac 1{\beta-1},\frac w{\beta-1})$ is the translation by $Q_2$ (a point of order $3$), and thus for any rational point $P$ on $E_1$, exactly one of $\{P,\tau(P),\tau^2(P)\}$ has a rational preimage in $E_3$ (has $F(\beta)^3$ rational).

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.