<s>[INST] Applying the quadratic Tschirnhausen transformation As per my previous question, I attempted to take dxiv's approach, though I can't seem to make much headway. Considering the simpler problem $x^3=x+a$ and the substitution $y=x^2+mx+n$, I got the following:
\begin{align}y^2&=x^4+2mx^3+(m^2+2n)x^2+2mnx+n^2\\&=(m^2+2n+1)x^2+(2mn+2m+a)x+(n^2+2am)\\&=\dot\ell x^2+\dot mx+\dot n\end{align}
\begin{align}y^3&=(m^2+2n+1)x^4+(2mn+2m+a)x^3+(n^2+2am)x^2\\&~~~~~+(m^3+2mn+m)x^3+(2m^2n+2m^2+am)x^2+(mn^2+2am^2)x\\&~~~~~+(m^2n+2n^2+n)x^2+(2mn^2+2mn+an)x+(n^3+2amn)\\&=(m^2+2n+1)x^4+(m^3+4mn+3m+a)x^3\\&~~~~~+(3m^2n+2m^2+3n^2+3am+n)x^2\\&~~~~~+(3mn^2+2am^2+2mn+an)x\\&~~~~~+(n^3+2amn)\\&=(3m^2n+3m^2+3n^2+3am+3n+1)x^2\\&~~~~~+(3mn^2+m^3+3am^2+6mn+3m+3an+2a)x\\&~~~~~+(am^3+n^3+6amn+3am+a^2)\\&=\ddot\ell x^2+\ddot mx+\ddot n\end{align}
Finding suitable coefficients for a cubic in $y$ leads me to
$$(2m-m^3+a-1)y^3+uy^2+vy=w\implies$$
\begin{cases}u=\ddot\ell m-\ddot m=3m^3n+3am^2-3mn-2m-3an-2a\\v=\dot\ell^2m-\ddot\ell\dot m-\dot\ell\ddot\ell m-\dot\ell\ddot m=\dots\end{cases}
which seems to leave me with quartics and possibly worse, when trying to set $u=v=0$.
Do I really need to expand that $v$? Have I made a mistake? Or is this simply a terrible path to go down and I should try using Newton's identities, as explained here?
 [/INST] I can follow the calculations all the way down to the last cubic. But when equating the following determinant to $\,0\,$ ...
$$
\scriptsize{
\left|
\begin{matrix}
1 & m & n  - y\\
m^2+2n+1 & 2mn+2m+a & n^2+2am-y^2 \\
3m^2n+3m^2+3n^2+3am+3n+1 & 3mn^2+m^3+3am^2+6mn+3m+3an+2a & am^3+n^3+6amn+3am+a^2 - y^3
\end{matrix}
\right|
}
$$
... the coefficient I get for $\,y^2\,$ is $\,\color{blue}{u} = - (3 m^3 n \color{red}{+ 2 m^3} - 3 m n - 2 m - 3 a n - 2 a)\,$, instead of the posted $\,3m^3n\color{red}{+3am^2}-3mn-2m-3an-2a\,$. This $\,\color{blue}{u}\,$ factors as $\,(3 n + 2) ( - m^3 + a + m)\,$, which gives the immediate choice $\,n = -2/3\,$ to drop the $\,y^2\,$ term, then what's left is a quadratic in $\,m\,$ to drop the linear term next.
This is consistent with WA's collect[ resultant[ x^3 - x - a, y - x^2 - m x - n, x ], y ] $\small= -a^2 - a m^3 + y (-3 a m - m^2 + 3 n^2 + 4 n + 1) + 3 a m n + a m + m^2 n - n^3 - 2 n^2 + \color{red}{(-3 n - 2)} y^2 - n + y^3$.

[ EDIT ]   To simplify the calculations somewhat, one can multiply the first column of the determinant above by $\,m\,$ and subtract it from the second column, which gives:

$$
\scriptsize{
\left|
\begin{matrix}
1 & 0 & n  - y\\
m^2+2n+1 & -m^3+m+a & n^2+2am-y^2 \\
3m^2n+3m^2+3n^2+3am+3n+1 & - 3m^3n-2m^3+3mn+2m+3na+2a & am^3+n^3+6amn+3am+a^2 - y^3
\end{matrix}
\right| \\
= \left|
\begin{matrix}
1 & 0 & n  - y\\
m^2+2n+1 & -m^3+m+a & n^2+2am-y^2 \\
3m^2n+3m^2+3n^2+3am+3n+1 & (- m^3 + m + a)(3 n + 2) & am^3+n^3+6amn+3am+a^2 - y^3
\end{matrix}
\right| \\
= -(m^3-m-a) \left|
\begin{matrix}
1 & 0 & n  - y\\
m^2+2n+1 & 1 & n^2+2am-y^2 \\
3m^2n+3m^2+3n^2+3am+3n+1 & 3 n + 2 & am^3+n^3+6amn+3am+a^2 - y^3
\end{matrix}
\right|
}
$$
The latter determinant is a monic cubic in $y$ which (of course) matches WA's resultant.
</s>