I. Methods
Using a quartic Tschirnhaus transformation, one can eliminate three terms $x^{n-1}, x^{n-2}, x^{n-3}$ without solving a $1\times2\times3 = 6$-deg equation, a simple explanation of which is given in this post.
However, using a quintic Tschirnhaus transformation, one can eliminate three terms $x^{n-1}, x^{n-3}, x^{n-5}$ without solving a $1\times3\times5 = 15$-deg equation. For the case of the sextic, this reduces it to the "cubic" $z^6+Az^4+Bz^2+C = 0$.
Like the first method, the second consist of two steps.
II. Step 1
Using the general sextic,
$$x^6+a_1x^5+a_2x^4+a_3x^3+a_4x^2+a_5x+a_6=0\tag1$$
First, get rid of the $(x^5,x^3)$ terms using a quadratic Tschirnhausen transformation,
$$x^2+mx+n-y=0\tag2$$
by eliminating $x$ between $(1)$ and $(2)$ using resultants and collecting the new variable $y$ to get,
$$y^6+b_1y^5+b_2y^4+b_3y^3+b_4y^2+b_5y+b_6=0\tag{3a}$$
Set $b_1 = b_3=0$ and solve for $(m,n)$ which will require at most only a cubic. Thus, the partially reduced,
$$y^6+b_2y^4+b_4y^2+b_5y+b_6=0\tag{3b}$$
The next step is to get rid of $b_5$.
III. Step 2
Use a quintic Tschirnhausen transformation,
$$y^5+ay^4+by^3+cy^2+dy+e-z=0\tag4$$
and eliminate $y$ between $(3b)$ and $(4)$ to get,
$$z^6+c_1z^5+c_2z^4+c_3z^3+c_4z^2+c_5z+c_6=0\tag5$$
First, use the variable $e$ to set $c_1 = 0$. Second, the coefficient of $z^3$ has form,
$$c_3:=P_0d^3+P_1d^2+P_2d+P_3 = 0$$
But because of Step 1, then actually $P_0=0,$ hence $c_3$ is only a quadratic in $d$. The three unknowns $(a,b,c)$ can then suffice to solve the three equations $P_1 = P_2 = P_3 = 0,$ making $d$ a "free" parameter. We can then use it to get rid of the $z$ term,
$$c_5:=Q_0d^5+Q_1d^4+Q_2d^3+Q_3d^2+Q_4d+Q_5 = 0$$
a quintic in $d$. Since $c_1 = c_3 = c_5 = 0,$ so the general sextic becomes a "cubic" in $z^2$,
$$z^6+c_2z^4+c_4z^2+c_6 = 0\tag6$$
This method is easily generalized to remove the three terms $x^{n-1}, x^{n-3}, x^{n-5}$ from higher equations.
IV. Sextic resolvent of the sextic
This method comes so close to what Jerrard wanted, namely "to solve the general sextic by solving a quintic" (just like a quartic is solved by a cubic). Unfortunately $(a,b,c)$ are roots of $1\times2\times3 = \text{deg-6}$ equations, and there seems to be no way of reducing the degree. (Joubert and Valentiner also had sextic resolvents in their methods.)
V. Questions about the sextic and nonic
Of course, the next step is remove four terms. Hilbert, using a geometric approach involving 27 straight lines, managed to remove four terms from the nonic to reduce it to the form,
$$x^9+Ax^4+Bx^3+Cx^2+Dx+1 = 0$$
Recall that in Hilbert's 13th problem, he was investigating the similar,
$$x^7+Ax^3+Bx^2+Cx+1 = 0$$
In Sutherland's 2021 paper, the impression I got (from Section 4) was that Hilbert's transformation was not in radicals and a quintic was involved. However, in Heberle's 2021 paper (p.8) it seems one can set coefficients $A_1 = A_2 = \dots = A_k = 0$ "without solving any equation of degree higher than $k$," so Hilbert's was just $k=4$. More details can also be found in Wolfson's 2020 paper. (Seems there are several papers on Tschirnhausen transformation all of a sudden, which is nice.)
So some questions are:
- Is there really no way to reduce the degree of the sextic's $(a,b,c)$ so only quintics are involved?
- Was Hilbert's nonic transformation and its coefficients in radicals? Or were only the coefficients in radicals? Or were neither transformation nor its coefficients in radicals?*
