I. Sextic
The general sextic can be reduced to a two-parameter form,
$$Ax^6+Bx^2+x+1=0$$ $$y^6+y^2+ay+b=0$$
To transform $P(x)\to P(y)$ and vice versa is easy. They have solutions,
$$x = -\sum_{j=0}^\infty \sum_{k=0}^\infty (+1)^k \frac{(2j+6k)!}{j!k! (j + 5 k + 1)!}\; A^j B^k$$
and,
$$y_k=\sum_{n=1}^\infty\frac{\left(\frac n6\right)!}{n!\,\Gamma\big(2-\frac{5n}6\big)}\,e^\frac{(2k+1)\pi i n}6(-p)^{1-\frac{5n}6}(-q)^\frac n6\,{_2 F_1}\Big(1-n,-\frac n6;2-\frac{5n}6;\frac pq\Big)$$
where $p,q=\frac{-a\pm\sqrt{a^2-4b}}2.$ The first is by Robert Israel, while the second is by Tyma Gaidash in this post. Both are limited by different radii of convergence.
However, I was surprised by the second solution since the hypergeometric function ${_2F_1}$ is a one-variable function, whereas what should solve the general sextic is the Kampé de Fériet function which is two-variable. This has implications for deg-$7$ equations.
II. Septic
The general septic can be reduced to the 3-parameter,
$$u^7+Au^3+Bu^2+Cu+1=0$$
By a minor change of variables, this can be expressed as,
$$v^7+(v+p)(v+q)(v+r)=0$$
The Lauricella series is a three-variable function and can perhaps solve the reduced septic. But Hilbert's 13th Problem asks if the septic's root can be expressed as the composition of a finite number of two-variable functions.
III. Question
Which brings me to my question. Can we solve the reduced septic in the same manner as Tyma did for the reduced sextic? Would the one-variable ${_2F_1}$ still do, or do we need something more general?
Reduced Septic: $\def\F{\operatorname F}$
$\def\FD{\operatorname F_\text D}$
We apply Lagrange reversion like in this post:
$$z^7+(z+p)(z+q)(z+r)=0\iff z=e^\frac{(2k+1)\pi i }7(z+p)^\frac17 (z+q)^\frac17 (z+r)^\frac17\implies z_k=\sum_{n=0}^\infty \frac{e^\frac{(2k+1)\pi i (n+1)}7}{(n+1)!}\left.\frac{d^n}{dz^n}(z+p)^\frac{n+1}7 (z+q)^\frac{n+1}7 (z+r)^\frac{n+1}7\right|_0$$
Now use a strategy like in this arXiv paper with the multinomial product rule and Pochhammer symbol $(m)_n$:
$$\left.\frac{d^n}{dz^n}(z+p)^\frac{n+1}7 (z+q)^\frac{n+1}7 (z+r)^\frac{n+1}7\right|_0=\sum_{n_1+n_2+n_3=n}\binom n{n_1,n_2,n_3}(pqr)^\frac{n+1}7\left(-\frac1p\right)^{n_1} \left(-\frac1q\right)^{n_2} \left(-\frac1r\right)^{n_3}\left(-\frac{n+1}7\right)_{n_1} \left(-\frac{n+1}7\right)_{n_2} \left(-\frac{n+1}7\right)_{n_3}$$
where the generalized Chu-Vandermonde identity $(1)$ and Lauricella $\F_\text D ^{(n)}$ function transformation $(10)$ in the linked paper appear. We have $3$ indices $n_j$, so we use $\F^{(2)}_\text D$ which reduces to the two-variable Appell $\F_1$,
$$F_1\big(a;b_1,b_2;c;\beta_1,\beta_2 \big)$$
With factorial power $n^{(m)}$, we get:
$$\bbox[2.5px,border:5px groove blue]{\begin{align} &z^7+(z+p)(z+q)(z+r)=0\implies \\ &z_k =\sum_{n=1}^\infty e^\frac{(2k+1)\pi i n}7\frac{(pq)^\frac n7}{r^{\frac{6n}7-1}n!}\left(\frac n7\right)^{(n-1)}\F_1\left(1-n;-\frac n7,-\frac n7;2-\frac{6n}7;\frac rp,\frac rq\right)\\ &\text{for}\,k=0,\dots, 6.\end{align}}$$
Plugging in arbitrary parameters $(p,q,r)$ gives all roots which match the actual ones. Alternatively, using the more familiar gamma function $\Gamma(n)$ and by permuting $(p,r)$,
$$z_k =\sum_{n=1}^\infty e^\frac{(2k+1)\pi i n}7 \, \frac{(qr)^\frac n7}{p^{\frac{6n}7-1}} \, \frac{\;\left(\frac n7\right)!}{n!\,\Gamma\big(2-\frac{6n}7\big)}\, \F_1\left(1-n;-\frac n7,-\frac n7;2-\frac{6n}7;\frac pq,\frac pr\right)$$
makes it very similar to hypergeometric ${_2F_1}$ solution to the reduced sextic in this post,
$$z_k=\sum_{n=1}^\infty e^\frac{(2k+1)\pi i n}6 \frac{(-q)^\frac n6}{(-p)^{\frac{5n}6-1}} \frac{\left(\frac n6\right)!}{n!\,\Gamma\big(2-\frac{5n}6\big)}\,{_2 F_1}\left(1-n,-\frac n6;2-\frac{5n}6;\frac pq\right)\qquad$$
This same method works on the reduced octic, but for degrees $m\geq9$, we cannot factor part of the polynomial in radicals. For example $z^9+a_5z^5+\dots +a_1=z^9+(z-r_1)\dots(z-r_5)=0$ has $r_j$ that cannot be put in terms of radicals in general. This means the generalized Chu-Vandermonde identity cannot be used to find a single sum of a Lauricella function solution for its roots $z_k$, like in this answer, unless we allow “non-radical” $r_j$.
Reduced Octic:
A sum of the Lauricella $\FD^{(3)}$ function appears after continuing the pattern:
$$\bbox[2.5px,border:5px groove blue]{\begin{align}&z^8+(z+p)(z+q)(z+r)(z+s)=0\implies \\ &z_k =\sum_{n=1}^\infty e^\frac{(2k+1)\pi i n}8\frac{(qrs)^\frac n8\left(\frac n8\right)!}{p^{\frac{7n}8-1}\left(1-\frac{7n}8\right)!n!}\FD^{(3)}\left(1-n;-\frac n8,-\frac n8,-\frac n8;2-\frac{7n}8;\frac pq,\frac pr,\frac ps\right)\\ &\text{for}\,k=0,\dots, 7.\end{align}}$$
shown here using $\FD^{(3)}$’s triple sum definition.