Define the Ramanujan theta function $f(a,b)$ as,
$$f(a,b) = \sum_{n=-\infty}^{\infty} a^{n(n+1)/2}\,b^{n(n-1)/2}$$
and the Dedekind eta function,
$$\eta(\tau) = q^{1/24}\prod_{n=1}^{\infty} \left(1-q^n\right)$$
A. In Ramanujan's Notebooks III, p. 372, Entry 8 he gives a series of identities which can be summarized as follows: let $q = e^{2\pi i \tau}$ and
$$h_k = (-1)^{k-1}q^{k(-13+3k)/26}\,\frac{f(-q^{2k},-q^{13-2k})}{f(-q^{k},-q^{13-k})} \;\;\text{for}\;k = 1,\dots,6.$$
Then the 3 roots of the cubic
$$\left(\frac{\eta(\tau)}{\eta(13\tau)}\right)^2=\frac{y^3+y^2-4y+1}{y(1-y)}\tag{0}$$
are
$$y_1 = h_1h_5,\;\;y_2 = h_2h_3,\;\;y_3 = h_4h_6.$$
B. I suspected this might have an analogue for $n=25$, the last integer such that $n-1$ divides $24$. Define
$$h_k = (-1)^{k-1}q^{k(-25+3k)/50}\,\frac{f(-q^{2k},-q^{25-2k})}{f(-q^{k},-q^{25-k})}\;\;\text{for}\;k = 1,\dots,12.\tag{1}$$
One pair is a constant, $h_5 h_{10} = -1$. After some experimentation with Mathematica’s integer relations, the other five pairs were the roots of a quintic which, to my surprise, turned out to be the Emma Lehmer quintic for real cyclic fields,
$$1+(10+10v+4v^2+v^3)x+(5+15v+11v^2+5v^3+v^4)x^2\\ -2(5+5v+3v^2+v^3)x^3+v^2x^4+x^5=0.\tag{2}$$
This is solvable in radicals for any $v$. Moreover, if we let
$$v=\frac{1}{R(q^5)}-R(q^5)-1= \frac{\eta(\tau)}{\eta(25\tau)},\tag{3}$$
where $R(q)$ is the Rogers-Ramanujan continued fraction, then it can be empirically observed that the five roots are
$$ x_1,\, \dots,\, x_5 = h_1 h_7,\; h_2 h_{11},\; h_3 h_4,\; h_6 h_8,\; h_9 h_{12},\tag{4} $$
hence Emma Lehmer's quintic is analogous to Ramanujan's cubic.
Question: Anybody knows how to prove that if $v$ is defined by $(3)$, then these $x_i$ are indeed the roots?