**I. Degree 8**


Assume the $j\_i$ to be free parameters. The following octics in $x$ belong to $8T43,$ have group $\text{PGL}(2,7)$, and order $2\times168 = 336.$


\begin{align}
{j\_1}\; &=\frac{(x^2 + 5x + 1)^3(x^2 + 13x + 49)}x\\
{j\_2}^2 &=\frac{j\_2\,(-7x^4 - 196x^3 - 1666x^2 - 3860x + 49)+(x^2 + 14 x + 21)^4}x\\
{j\_3}\; &=\frac{(x + 1)^6(x^2 + x + 7)}x\\
{j\_4}^2 &=\frac{7 j\_4\,(x + 1)^4+(x + 1)^7 (x - 7)}x
\end{align}


Expanded out, some of the coefficients are quadratic in $j\_i$. They also have nice discriminants $D\_i$. (Note: $D\_2$ has another square factor I've suppressed for brevity. I have a feeling the second octic may have a simpler version.)


\begin{align}
D\_1 &= -7^7(j\_1-1728)^4\,{j\_1}^4\\
D\_2 &= -7^7(j\_2-256)^4\,{j\_2}^6 \\
D\_3 &= -7^7(j\_3-108)^3\,{j\_3}^5\\ 
D\_4 &= -7^7(j\_4-64)^4\,{j\_4}^{12}
\end{align}


For general $j\_i$, these octics are not solvable in radicals. However, if we substitute the following integers,


\begin{align}
j\_1 &= -640320^3\quad \\
j\_2 &= -63^2,\, 396^4\quad \\
j\_3 &= -300^3\quad \\
j\_4 &= -2^9,\, 2^{12}\quad
\end{align}


then they *become solvable*. The first two $j\_i$ are famous being in the Chudnovsky and Ramanujan pi formulas, and the last two can be used in [Ramanujan-Sato pi formulas](https://en.wikipedia.org/wiki/Ramanujan%E2%80%93Sato_series) as well.




---


**II. Eta quotients**


There are infinitely many such radical $j\_i$ given by,


\begin{align}
\quad j\_1 &=\left(\left(\frac{\eta(\tau)}{\eta(2\tau)}\right)^{8}+2^8 \left(\frac{\eta(2\tau)}{\eta(\tau)}\right)^{16}\right)^3 \\
\quad j\_{2} &=\left(\left(\frac{\eta(\tau)}{\eta(2\tau)}\right)^{12}+2^6 \left(\frac{\eta(2\tau)}{\eta(\tau)}\right)^{12}\right)^2 \\
\quad j\_{3} &=\left(\left(\frac{\eta(\tau)}{\eta(3\tau)}\right)^{6}+3^3 \left(\frac{\eta(3\tau)}{\eta(\tau)}\right)^{6}\right)^2 \\ 
\quad j\_{4} &=\left(\left(\frac{\eta(\tau)}{\eta(4\tau)}\right)^{4} + 4^2 \left(\frac{\eta(4\tau)}{\eta(\tau)}\right)^{4}\right)^2 = \left(\frac{\eta^2(2\tau)}{\eta(\tau)\,\eta(4\tau)} \right)^{24} 
\end{align}


where $\eta(\tau)$ is the [*Dedekind eta function*](https://en.wikipedia.org/wiki/Dedekind_eta_function#Eta_quotients) and $j\_1$ is just the j-function. (For examples of $\tau$, click on the link.)


Thus, the octics in $x$ are also formulas for the $j\_i$ and where $x$ involve eta quotients (though I didn't include them to prevent clutter).




---


**III. Degree 6**


So far, I've only found two with group $\text{PGL}(2,5)$ hence $6T14$ and order $2\times60 = 120$. (***Edit***: For instant comparison, we add the other two $6T14$ from Joachim König's answer, in red.)


\begin{align}
j\_1 &=\frac{(x^2 + 10x + 5)^3}x\\
\color{red}{j\_2} &=\frac{(x+1)^4(x^2+6x+25)}{x}\\
\color{red}{{j\_3}^2} &= \frac{j\_3\,(2 y^6 + 29y^5 + 85y^4 + 50y^3) + 5^4 y^6}{(2y - 1)} \\
j\_4 &=\frac{(x + 1)^5 (x + 5)}x
\end{align}


with discriminants,


\begin{align}
D\_1 &= 5^5\,(j\_1-1728)^2\,{j\_1}^4\\
D\_2 &= 5^5\,(j\_2-256)^3\,{j\_2}^3 \\
D\_3 &= 5^5(j\_3-108)^3\,{j\_3}^{11}\\ 
D\_4 &= 5^5\,(j\_4-64)^2\,{j\_4}^4
\end{align}


It is preferred the constant factor is only $5^5$. *Note*: $D\_3$ actually has the extra square factor $(32 j\_3+84375)^2$, but I truncated it for aesthetics.




---


**IV. Question:**


So are there analogous sextic $6T14$ formulas in $x$ for $j\_2, j\_3$, where the $j\_i$ are linear or quadratic, and discriminants similar to the others? (Note: Has been answered in the affirmative.)




---


**V. Addendum:**


*(The addendum has been moved as an "answer" to prevent clutter and loading issues.)*



Finding a sextic analogue to the solvable octic $\frac{(x + 1)^6(x^2 + x + 7)}x = -k^3$ where $e^{(\pi/3)\sqrt{d}}\approx k^3+41.999999999999\dots$
Didn't think about eta quotients, but if the suggested analogies of ramification types are anything to go by, then
$$j\_2 = \frac{(x+1)^4(x^2+6x+25)}{x}$$
(which matches the prescribed discriminant exactly), and
$$j\_3^2 = \frac{2(x+1)^3(x^3+32x^2+231x+400) j\_3 + 5^4(x+1)^6}{64x}$$
(or possibly some reparametrization of it). The last equation has additional divisors a $2$-power as well as a square of a linear factor (however not corresponding to a branch point of the function field extension); on the other hand, the quintic subextension of the Galois closure could be parameterized by $j\_3 =y^3(y-5)^2$, which does give exactly matching discriminant. Not sure if there's any direct reason why reparametrization should allow for extra factors to vanish, just like for general number fields it's not always possible to find a polynomial whose discriminant matches the field discriminant.


