The octic Ramanujan-Selberg continued fraction $S(q)$ and $x^8+y^8=1$ can solve the Bring quintic. So can the Rogers-Ramanujan continued fraction $R(q)$ and $x^5+y^5=1.\,$ It turns out Ramanujan's cubic continued fraction $C(q)$ and $x^3+y^3=1$ can do so as well. Thus, all three Platonic symmetries (octahedral, icosahedral, and tetrahedral) can be used to solve the quintic, answering this MO post about the Monster in the affirmative.
I. Cubic continued fraction
Let $q=e^{2\pi i \tau}$ and Dedekind eta function $\eta(\tau)$, then,
$$C(\tau) = \cfrac{q^{1/3}}{1 + \cfrac{q+q^2}{1 + \cfrac{q^2+q^4}{1 + \ddots}}} = \frac{\eta(\tau)\,\eta^3(6\tau)}{\eta(2\tau)\,\eta^3(3\tau)} $$
Also define a function needed later as $D(\tau) = \dfrac{\eta(\tau)\,\eta(2\tau)}{\eta(3\tau)\,\eta(6\tau)}$ such that,
$$\frac1{C^3(\tau)}-8C^3(\tau) = D^4(\tau)+7\quad$$
Note that $\dfrac1{C^3(\tau)}$ and $D^4(\tau)$ are both McKay-Thompson series for the Monster, where the former is class 6E (A105559) and the latter is class 6D (A121667).
II. Modular Equations
The modular equation between $C(\tau)$ and $C(3\tau)$ can solve the Fermat cubic $x^3+y^3=1$,
$$\left(\frac{3}{\alpha+3}\right)^3+\left(\frac{\beta}{\beta+9}\right)^3=1$$
\begin{align} \alpha &= 4C^2(\tau)\,+\,\frac1{C(\tau)}\,-3\, = \left(\frac{\eta(\tau/3)}{\eta(3\tau)}\right)^3\\ \beta &= 4C^2(3\tau)+\frac1{C(3\tau)}-3 = \left(\frac{\eta(\tau)}{\eta(9\tau)}\right)^3 \end{align}
The modular equation between $C(\tau)$ and $C(5\tau)$ is a bit long (I haven't yet found a neat expression),
$$u^6 - u v + 5u^4v + 5u^2v^2 - 10u^5v^2 - 20u^3v^3 + 5u v^4 + 20u^4v^4 - 10u^2v^5 - 16u^5v^5 + v^6 = 0$$
If $v = C(\tau)$, then the six roots $u$ are,
$$u = C(5\tau),\;C\big(\tau/5\big),\;C\big(\tfrac{\tau+3}{5}\big),\;C\big(\tfrac{\tau+6}{5}\big),\;C\big(\tfrac{\tau+9}{5}\big),\;C\big(\tfrac{\tau+12}{5}\big)$$
Define,
$$H(\tau) = \left(\frac{C(5\tau)-C(\tau/5)}{\sqrt5\, C^3(\tau)\,D^3(\tau)}\right) \left(C\big(\tfrac{\tau+3}{5}\big)-C\big(\tfrac{\tau+12}{5}\big)\right) \left(C\big(\tfrac{\tau+6}{5}\big)-C\big(\tfrac{\tau+9}{5}\big)\right)$$
Expanding
$$\prod_{n=0}^4\big(x-H(\tau+4n)\big)=x^5-5x-\frac{D^4(\tau)+27}{D^3(\tau)}=0$$
Equating this to a generic Bring quintic $x^5-5x-d=0$, one should solve for $\tau$ in,
$$\frac{D^4(\tau)+27}{D^3(\tau)}=d$$
which can be done after a lot of math.
III. Solution
The Bring quintic
$$x^5-5x-d = 0$$
has solution for $n=(0,1,2,3,4)$,
$$x_n = H(\tau+4n)$$
where $H(\tau)$ is defined by the cubic continued fraction $C(\tau)$ above and,
$$\quad \tau =\frac{K'(k)}{K(k)}\sqrt{-1} \,=\, \frac{_2F_1\big(\tfrac12,\tfrac12,1,\,1-k^2\big)}{_2F_1\big(\tfrac12,\tfrac12,1,k^2\big)}\sqrt{-1}$$
with $k$ as any appropriate root of the octic,
$$\;256 - 512k^2 + (384 - d^4)k^4 - (128-d^4)k^6 + 16k^8 = 0$$
Note: Compare to the octic of the previous three methods,
$$256k^8 - 512k^6 + (384 + c^4)k^4 - (128 + c^4)k^2 + 16\, = \,0$$
If $d\to(-1)^{1/4}c,\,$ and $k\to1/k,\,$ then it is essentially the same equation.
IV. Question
Summary of the 8 methods using elliptic modulus $k$ are,
$$\frac{\sqrt2\,\eta\big(\tfrac{\tau}2\big)\,\eta^2(2\tau)}{\eta^3(\tau)}\quad\text{and}\quad\frac{\sqrt2\,\eta(2\tau)}{\;\eta(\tau/2)}\quad\quad $$ $$\vartheta_4(0,z)\quad\text{and}\quad\vartheta_3(0,z)$$ $$g(\tau)\quad\text{and}\quad G(\tau)$$ $$R(q)\quad\text{and}\quad C(q)$$
with the first pair in this post, the Jacobi theta functions here, the Ramanujan $g$- and $G$-functions in this post, and the Rogers-Ramanujan continued fraction $R(q)$ here.
Q: What's the big picture why solving for $\tau$ in vastly different equations like,
\begin{align} c &= \sqrt8\,\frac{2g^8(\tau)-G^8(\tau)}{G^2(\tau)}\\[6pt] d &= \frac{D^4(\tau)+27}{D^3(\tau)} \end{align}
then (after a lot of algebraic manipulation) the same quartic or octic in $k$ keeps popping up, even for the cubic continued fraction $C(q)$?
