How to solve this equation in radicals?

Question

How to solve the equation $x^6-2\varphi^5x^5+2\varphi x+\varphi^6=0$ in radicals? where $\varphi = \phi^{1/4}$ and $\phi$ is the golden ratio.

Answer 1

The Schlaefli modular equation, $$\bigg(\frac{u}{v}\bigg)^3+\bigg(\frac{v}{u}\bigg)^3=2 \bigg(u^2v^2-\frac{1}{u^2v^2}\bigg)$$ or expanded out as a sextic,

$$u^6 - 2 u^5 v^5 + 2 u v + v^6 = 0\tag1$$

has the closed-form solution, $$u = G_{25n},\quad v = G_n$$ with Ramanujan G-function $G_n$. For rational $n>0$, then $u,v$ are radicals. Courtesy of an answer by G. Manco, it turns out $(1)$ belongs to that special class of sextics that can be solved by quintics (similar to how quartics can be solved by cubics).

Given $v = G_n$ in $(1)$, an alternative closed-form solution for $u$ is then,

$$u=G_{25n}= \frac{(G_n)^5}{10}\Big((\beta+1)(x+1)^2-5\Big)\tag2$$

where $x$ is the real root of the solvable DeMoivre quintic, $$x^5-5\alpha x^3+5\alpha ^2 x -\alpha (\alpha ^2-2\alpha +8)=0$$ and, $$\alpha =\frac{2\,G_{n}}{(G_{n/25})^5},\quad \beta= \frac{2\,G_{n/25}}{(G_n)^5}$$ For example, knowing just $G_1 = 1,\; G_{1/5}=G_5 = \phi^{1/4},\; G_{25} = \phi$, then $(2)$ is an iterative method to express in radicals all $G_n$ with $n=5^m$ in terms of the golden ratio $\phi$.

Answer 2

G125 is here!I used a 25th order modular equation (Notebook n.2 ch. xix p.231