Given the Ramanujan theta function $f(a,b)$ and the Rogers-Selberg identities,
\begin{align} U_1 &= \frac{f(-q,-q^6)}{f(-q^2)} = \sum_{n=0}^\infty \frac {q^{2n^2+2n}} {(q^2;q^2)_n\,(-q;q)_{2n+1}} = \prod_{n=1}^\infty (1-q^{7n-1})(1-q^{7n-6})\,x_7\\ U_2 &= \frac{f(-q^2,-q^5)}{f(-q^2)} = \sum_{n=0}^\infty \frac {q^{2n^2+2n}} {(q^2;q^2)_n\,(-q;q)_{2n}} \,=\, \prod_{n=1}^\infty (1-q^{7n-2})(1-q^{7n-5})\,x_7\\ U_3 &= \frac{f(-q^3,-q^4)}{f(-q^2)} = \sum_{n=0}^\infty \frac {q^{2n^2}} {(q^2;q^2)_n\,(-q;q)_{2n}} \,=\, \prod_{n=1}^\infty (1-q^{7n-3})(1-q^{7n-4})\,x_7 \end{align}
where $x_7 = \frac{(1-q^{7n})}{(1-q^{2n})}$. (Caveat: I've reversed the naming convention in Mathworld since this seems more "natural".) First let,
\begin{align} a &= -q^{61/168}\,U_1\\ b &= \; q^{13/168}\;U_2\\ c &= q^{-11/168}\,U_3 \end{align}
such that,
$$a^3b+b^3c+c^3a = 0$$
so the integer $168$ reflects the order of the Klein quartic curve above. Then given the Dedekind eta function $\eta(\tau)$, define,
\begin{align} V_1 &= \frac{\eta(7\tau)}{\eta(\tau)} \left(\frac{\eta(2\tau)}{\eta(7\tau)}\right)^3 \left(-q^{61/168}\,U_1\right)^3\\ V_2 &= \frac{\eta(7\tau)}{\eta(\tau)} \left(\frac{\eta(2\tau)}{\eta(7\tau)}\right)^3 \left(q^{13/168}\,U_2\right)^3\\ V_3 &= \frac{\eta(7\tau)}{\eta(\tau)} \left(\frac{\eta(2\tau)}{\eta(7\tau)}\right)^3 \left(q^{-11/168}\,U_3\right)^3 \end{align}
Then the cubic formed by their $7$th powers
$$P(u) = (u-V_1^7)(u-V_2^7)(u-V_3^7) = 0$$
has coefficients in the eta quotient $ n = \left(\frac{\eta(\tau)}{\eta(7\tau)}\right)^4,$
$$u^3 - (1369 + 854 n + 207 n^2 + 23 n^3 + n^4) u^2 + (17839 + 12082 n + 3343 n^2 + 472 n^3 + 34 n^4 + n^5) u + 1 =0$$
Note that for $\tau = \sqrt{-d}$, since $n$ and other eta quotients are radicals, then this cubic proves that $q^{61/168}\,U_1,\, q^{13/168}\,U_2,\,q^{-11/168}\,U_3$ are also radicals. (Just like their quintic counterparts $q^{-1/60}G(q)$ and $q^{11/60}H(q)$ where the integer $60$ reflects the order of the icosahedron.)
We then form the septic,
$$P(y)=\prod_{k=0}^6 \Big(y - (\zeta^k\, V_1 + \zeta^{4k}\,V_2 + \zeta^{2k}\,V_3)\Big) = 0$$
with $\zeta = e^{2\pi i/7}$ which also has coefficients in $n$,
$$y^7 + 14 y^4 - 21 y^3 - 14 (46 + 13 n + n^2) y^2 - 7 (249 + 114 n + 18 n^2 + n^3) y - (1348 + 854 n + 207 n^2 + 23 n^3 + n^4) = 0$$
Just like in the first family, a root of $P(y)$ is then,
$$y = u_1^{1/7} + u_2^{1/7} + u_3^{1/7}$$
and is now solvable in radicals for any $n$.
Question: So are there infinitely many parametric families of septics (solvable by irreducible cubics) we can construct in a similar manner, or not many?
