On solvable octic trinomials like $x^8-5x-5=0$
Solution 1:
Using brute force approach I have found few others, such as $$x^8+9x+9$$ $$x^8+64x+112$$ $$x^8+75x+150$$
See https://sites.google.com/site/klajok/polynomials/x8-plus-ax-plus-b-is-0
$\color{green}{Added:}$
Given pairs of rational numbers $\left(\alpha,\beta\right)$ such that $2\alpha^2+6\alpha+1=\beta^2$. Define the following parameters:
$$u=\frac{2\alpha+1-\beta}{4},\quad v=\frac{1-\beta}{8},\quad w=\frac{\alpha}{8}\left(3\alpha-2\beta+3\right)$$ $$A=\frac{\alpha u}{2}\left(\alpha+1-4u\right),\quad B=w^2 - \alpha v^2$$
then the following identity is satisfied: $$x^8+Ax+B =\\ \left(x^4+\sqrt{\alpha}x^3+\left(-\frac{\sqrt{\alpha}}{2}+\frac{\alpha}{2}\right)x^2+\left(u\sqrt{\alpha}-\frac{\alpha}{2}\right)x+\left(v\sqrt{\alpha}+w\right)\right)\\ \left(x^4-\sqrt{\alpha}x^3+\left(\frac{\sqrt{\alpha}}{2}+\frac{\alpha}{2}\right)x^2+\left(-u\sqrt{\alpha}-\frac{\alpha}{2}\right)x+\left(-v\sqrt{\alpha}+w\right)\right)$$
Let exclude the pairs $\left(\alpha, \beta \right) \in \left\{ \left(0, -1\right), \left(0, 1\right), \left(1, 3\right) \right\}$ for which trinomials degenerate to the simpler form where $AB=0$.
Observations: If $\sqrt{\alpha}$ is not a rational number then the corresponding octic trinomials are irreducible and solvable. Otherwise the trinomials are still solvable but they are not irreducible.
Notes:
All the $\left(\alpha, \beta \right)$ pairs can be easily enumerated: $$\left(\alpha, \beta \right) \in \left\{ \left( \frac{2q+6}{q^2-2}, \frac{q^2+6q+2}{q^2-2} \right) : q \in \mathbb Q \setminus \lbrace -3, 4 \rbrace \right\}$$ Excluded values $q=-3$ and $q=4$ correspond to the degenerate cases $(0, -1)$ and $(1, 3)$, respectively. The remaining degenerate case $(0,1)$ corresponds to the value $q$ at infinity (I found this "subparameterization" using the following Philip Gibbs' answer: https://math.stackexchange.com/q/1905148).
Examples and preliminary conjectures related to this parameterization is available under the same page: https://sites.google.com/site/klajok/polynomials/x8-plus-ax-plus-b-is-0 .
I searched for more examples of $C_2 \wr A_4$ and $C_2 \wr S_4$ for $x^8+ax+b$ and the only ones for integer $|a|,|b| \leq 100000000$ are:
$C_2 \wr A_4$: $$x^8+64x+112$$
$C_2 \wr S_4$: $$x^8+44x-33$$ $$x^8+768x+1344$$ $$x^8+31740x+113735$$ $$x^8+856251x-2023866$$ $$x^8+5992704x-304129728$$