Deriving Formulae for the roots of the quartic and cubic polynomials I have seen derivations of the general solution for the roots of fourth and third degree polynomials of 1 variable in Dummit & Foote's Abstract Algebra; however, it was by no means simple to me. I am curious to know if there is a more natural approach using techniques from Algebraic Geometry (i.e. using Groebner Bases and Elimination Theory).
My Attempt (cubic):
Consider the equation $f(x)=x^3 +ax^2 +bx+c$. By the Fundamental Theorem of Algebra, there exists $x_1, x_2, x_3 \in \mathbb{C}$ so that  $f(x) = (x-x_1)(x-x_2)(x-x_3)$. After multiplying everything out, we get the following system of polynomial equations by comparing coefficients.
\begin{cases} a=-x_1-x_2-x_3 \\ b = x_1 x_2 +x_1x_3+x_2x_3 \\ c= -x_1x_2x_3 \end{cases}
If we consider the ideal, $I = \langle x_1+x_2+x_3 +a, \;x_1x_2+x_1x_3+x_2x_3 - b,\; x_1x_2x_3 + c\rangle \subseteq \mathbb{C}[x_1,x_2,x_3]$.
 My idea is to find a Groebner basis using lexicographic order ($x_1 > x_2 > x_3$) which will hopefully lead to a simpler expression which can be solved. After putting the ideal into Macaulay 2, I arrived with the following basis for $I$.
$$\mathbb{G} = \{x_1+x_2+x_3 +a, \; x_2^2 + x_2x_3+x_3^2+ax_2+ax_3 +b, \; x_3^3 +ax_3^2 + bx_3 + c \}$$
However, this method doesn't seem to be leading anywhere since the final equation is exactly what we started with. 
My question: can this method can be modified so that it actually works, and if so, can it be extended for the quartic?
 A: 
However, this method doesn't seem to be leading anywhere since the final equation is exactly what we started with.

I'm pretty sure if you're going off of Vieta's formulas to try to solve the cubic in any way, you're just going to go in a circle. I don't know anything about algebraic geometry or Groebner bases, but I do know that the Vieta formulas lead to a system of equations and when you try to solve that system of equations, you get the original polynomial back. Therefore, trying to solve for the roots using the Vieta formulas is going to keep you going back in a circle.

For others trying to derive the cubic themselves, instead, think of ways you can simplify the cubic:


*

*Try getting rid of the $x$ term of the $x^2$ term. You can do this some substitution like $x=y-d$. If you make this substitution, for what value of $d$ does the coefficient of $x^2$ become $0$? For what value of $d$ does the coefficient of $x$ become $0$?

*Once you have gotten rid of one of the terms, you need to somehow reduce it to a linear or quadratic equation. However, you have an exponent of $3$ in this equation, so you'll likely need to make a linear or quadratic equation in terms of $x^3$. What kind of substitution do you think will lead to this?


These are hard questions to answer and I don't know if I'd be able to answer them if I didn't already know the solution, but good luck!
