Ramanujan's Class Invariant $G_{625}$ How to calculate the  Ramanujan Class Invariant $G_{625}$?
Equation is: 
$x^6-2\varphi^5x^5+2\varphi x+\varphi^6=0$.
$\varphi$ is the golden ratio.
 A: The Ramanujan $G_n$ and $g_n$ functions can be computed in Mathematica using the Dedekind eta function. Let $\tau=\sqrt{-n}$, then,
$$G_n=\frac{2^{-1/4}\,\eta^2(\tau)}{\eta\big(\tfrac{\tau}{2}\big)\,\eta(2\tau)}\quad \text{odd}\; n$$
$$g_n=\frac{2^{-1/4}\,\eta\big(\tfrac{\tau}{2}\big)}{\eta(\tau)}\quad \text{even}\; n$$
There is a modular equation between $u=G_{25n}$ and $v = G_{n}$ given by,
$$\bigg(\frac{u}{v}\bigg)^3+\bigg(\frac{v}{u}\bigg)^3=2 \bigg(u^2v^2-\frac{1}{u^2v^2}\bigg)$$
or expanded out,
$$u^6 - 2 u^5 v^5 + 2 u v + v^6 = 0\tag1$$

Your post seeks $u=G_{625}$ if we are given $v = G_{25} = \phi$ which is the golden ratio. The sextic factors nicely as
$$(u-1)\big(u^5 - 5 \phi^3  (u^4 + u^3 + u^2 + u) - \phi^6\big)=0$$
Thus, the problem is to solve that quintic. Define,
$$w_1=\left(6+4\phi+3\times 5^{1/4}\sqrt{\phi} \right)^{1/5}+\left(6+4\phi-3\times 5^{1/4}\sqrt{\phi} \right)^{1/5}$$
$$w_2 = \left(12\phi+4\times 5^{1/4}\sqrt{\phi^{-1}} \right)^{1/5} +\left(12\phi-4\times 5^{1/4}\sqrt{\phi^{-1}} \right)^{1/5} $$
then the solution to $(1)$,
$$u^6 - 2 u^5 v^5 + 2 u v + v^6 = 0\tag1$$
is given by,
$$\begin{align}
u &= G_{625}=\phi^3+\phi^2 (w_1+w_2) = 22.180323\dots\\
v &= G_{25} = \phi = 1.618033\dots
\end{align}$$
