<s>[INST] Is there a substitute that will eliminate terms of the degree $3$ and $2$ from the quartic polynomial? The standart form of the quartic equation:

$$ax^4+bx^3+cx^2+dx+e=0$$ where $$a\neq 0.$$

The purpose of the question is not to solve quartic. Resources related to the solution are sufficiently available. Just about special curiosity, I am looking for a special substitute that will transform the following:

$$ax^4+bx^3+cx^2+dx+e=0 \Longrightarrow z^4+pz+q=0$$
  where $$x=f(z)$$
Here $f(z)$ can be a linear or nonlinear complex function.

This substitute is known:

$$x\longmapsto x-\frac b{4a}$$

Which gives us
$$x^4+px^2+qx+r=0$$

Is a substitution known to eliminate term of degree $3$ and $2$  together?

 [/INST] We can assume that $a=1$ by dividing by it, so start with
$$ x^4 + bx^3 + cx^2 + dx + e = 0 $$
(there's no need for $f$, either).
We will use a more general substitution known as a Tschirnhaus substitution: put
$$ y = x^2 + \lambda x + \mu . $$
Then
$$ y^2 = x^4 + 2\lambda x^3 + 2(\mu+\lambda^2) x^2 + 2\lambda\mu x + \mu^2 = (2\lambda-b)x^3 + (\lambda^2+2\mu-c)x^2 + (2\lambda\mu-d)x + (\mu^2-e) \\
y^3 = \dotsb = (-b^3 + 2 b c - d + 3 b^2 \lambda - 3 c \lambda - 
 3 b \lambda^2 + \lambda^3 - 3 b \mu + 6 \lambda \mu) x^3 + (-b^2 c + c^2 + b d - e + 3 b c \lambda - 3 d \lambda - 
 3 c \lambda^2 - 3 c \mu + 3 \lambda^2 \mu + 3 \mu^2) + \dotsb \\
y^4 = \dotsb ,
 $$
and then since $1,y,y^2,y^3,y^4$ give five equations with four coefficients, we can find a linear combination so that they add up to zero, leaving us with the equation
$$ y^4 + (-b^2 + 2 c + b \lambda - 4 \mu) y^3 + (c^2 - 2 b d + 2 e - b c \lambda + 3 d \lambda + c \lambda^2 + 3 b^2 \mu - 6 c \mu - 3 b \lambda \mu + 6 \mu^2) y^2 + ()y + () = 0 , $$
where we're not interested in the last two coefficients. (This effectively calculates the resultant of the two polynomials $x^4+bx^3+cx^2+dx+e$ and $x^2+\lambda x-\mu-y$, which is generally best done with a computer.)
To eliminate the first two coefficients, we need
$$
-b^2 + 2 c + b \lambda - 4 \mu = 0 \\
c^2 - 2 b d + 2 e - b c \lambda + 3 d \lambda + c \lambda^2 + 3 b^2 \mu - 6 c \mu - 3 b \lambda \mu + 6 \mu^2 = 0 .
$$
The first of these is linear, the second quadratic, so it is easy to solve both simultaneously: we find
$$ \lambda = \frac{3 b^3 - 10 b c + 12 d \pm 2 \Delta}{3b^2-8c}, \qquad \mu = \frac{2 b^2 c - 8 c^2 + 6 b d \pm b \Delta}{6b^2-16c} , $$
where $\Delta^2 = 6 b^3 d - 28 b c d - 2 b^2 (c^2 - 6 e) + 4 (2 c^3 + 9 d^2 - 8 c e) $. I've left out a few of the details of the calculation, but they should be easy, if tedious, to fill in.
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