Is this system always solvable in radicals by quartics, octics, $12$-ics, etc?

Question

While considering this post, it made me wonder about its generalization in another direction and from the perspective of Galois theory.

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Question: Is it true that, given four constants ($\alpha,\beta,\gamma,\delta$), then the system, $$\begin{aligned} x_1^a+x_2^a+x_3^a+x_4^a &= \alpha\\ x_1^b+x_2^b+x_3^b+x_4^b &= \beta\\ x_1^c+x_2^c+x_3^c+x_4^c &= \gamma\\ x_1^d+x_2^d+x_3^d+x_4^d &= \delta \end{aligned}\tag1$$ can be solved in radicals $x_i$ for any positive integer power ($a,b,c,d$) and $a<b<c<d$?

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Example: Let's choose the year Back to the Future came out,

$$\begin{aligned} x_1+x_2+x_3+x_4 &= \color{brown}1\\ x_1^2+x_2^2+x_3^2+x_4^2 &= \color{brown}9\\ x_1^4+x_2^4+x_3^4+x_4^4 &= \color{brown}8\\ x_1^8+x_2^8+x_3^8+x_4^8 &= \color{brown}5 \end{aligned}$$

Two solutions are given by the roots of the octic,

$$4707 + 9036 x - 7495 x^2 - 5096 x^3 + 3024 x^4 + 1848 x^5 - 896 x^6 - 256 x^7 + 128 x^8 = 0$$

Magma says this is $8T47$, has order $1152=2^7\cdot3^2$ and is the largest solvable order for deg $8$. (In fact, it factors over $\sqrt{1441}$). Using the root $r_i$ numbering system of Mathematica, we find that,

$$x_1,\, x_2,\, x_3,\, x_4 = r_1,\, r_2,\, r_5,\, r_6\quad \text{(Solution 1)}\\ \text{or}\quad\quad\\ x_1,\, x_2,\, x_3,\, x_4 = r_3,\, r_4,\, r_7,\, r_8\quad \text{(Solution 2)}$$

Remarks: Using generic ($\alpha,\beta,\gamma,\delta$)

1. For exponents $a,b,c,d = 1,2,c,8$ and $c=3,4,5$, one ends up with a $8T47$ octic with order $1152=2^7\cdot3^2$.
2. For $a,b,c,d = 1,2,6,7$, one gets a $12T294$ with order $82944 = 2^{10} \cdot 3^4$.
3. For $a,b,c,d = 1,2,c,8$ and $c=6,7$, one now gets a $16T1947$ with order $7962624 = 2^{15}\cdot3^5$. These three are the largest solvable orders for their respective degrees.
4. Using others yields deg $20,24,$ etc.

So, is the system $(1)$ always solvable in radicals?

Answer 1

For $(a,b,c,d)=(1,2,3,20)$ and $(\alpha,\beta,\gamma,\delta)=(1,1,1,0)$ one computes that $x_1$ is a root of the irreducible polynomial $4 x^{20} - 20 x^{19} + 40 x^{18} - 40 x^{17} + 195 x^{16} - 704 x^{15} + 1050 x^{14} - 700 x^{13} + 475 x^{12} - 900 x^{11} + 900 x^{10} - 300 x^{9} + 130 x^{8} - 260 x^{7} + 130 x^{6} + 20 x^{4} - 20 x^{3} + 1$, which according to magma, has a non-solvable Galois group, namely the full wreath product $S_4^5\rtimes S_5$.

Without Magma: With the same values of $a,b,c,d,\alpha,\beta,\gamma,\delta$ as above, one computes (e.g. using Sage) that $x_1x_2x_3x_4$ is a root of $h(x)=4 x^{5} - 175 x^{4} + 300 x^{3} - 130 x^{2} + 20 x - 1$. Now $h(x-1)$ is an Eisenstein polynomial with respect to $5$, and modulo $3$ $h(x)=(x + 1) \cdot (x + 2) \cdot (x^{3} + 2 x^{2} + x + 1)$ is a factorization into irreducibles. Thus the Galois group of $h$ contains the alternating group $A_5$.