Lagrange studied the effects of permutations on values of resolvent. Permutations of roots in resolvents of cubic, quartic and quintic equations yield 2,3, and 6 distinct values. Hence the reduction approach that works for cubic and quartic equations wouldn't work for quintic. Permutations of roots of a polynomial, which is of great interest in Galois Theory.

As I understand, we solve quadratic equation $x^2+b \cdot x + c = 0$ (express $x_1$ and $x_2$) in two steps. We first find $-\frac{b}{2}$ corresponds to the axis of symmetry of a parabola. There is only one such point.

We note that because of symmetry $|-\frac{b}{2} - x_1| = |-\frac{b}{2} - x_2| = \sqrt{\frac{b^2}{4}-c}$. Hence, we can express roots as $x = -\frac{b}{2}\pm\sqrt{\frac{b^2}{4}-c}$ In a sence we fold parabola along the symmetry axis anmd reduce quadratic to a pair of first order equations. Is this interpretation correct?

What happens when we solve cubic equation? Depressed cubic has roots $x_1, x_2, x_3; x_1 + x_2 + x_3 = 0$. Cubic is solved by reduction to quadratic equation. Which to special points are we searching for? What are the three special points for quartic?


Depressed Cubic In depressed cubic $x_1+x_2+x_3 = 0$ because coefficient at $x^2$ is zero. Roots of a depressed cubic can be written as:
$x_1=m+n; x_2=m-n; x_3=-2\cdot m$, where $n= a+i\cdot b; a \cdot b=0; \{a,b,m\} \in R$

Here $m$ is always real and $n$ is either real or strictly imaginary. Lets look at these distinct cases. If a cubic has imaginary roots, then they split into a conjugate pair of complex roots and a real root. We then search for solution in the form $(x^2+r_1)\cdot(x+r_2)=0; \{r_1,r_2\} \in R$

Case of three real roots is more difficult. We cannot split them, because each root is real. Thus we restate the problem:

For a given depressed cubic find a right triangle ABC centered at origin such that vector sums $\{A+A_{vf}, B+B_{vf}, C+C_{vf}\}$ are roots of the cubic. Here $A_{vf}$ is the image of point A after a vertical flip. We can rotate and scale the triangle, hence, we have two degrees of freedom. Rotation of the triangle by $\frac{2 \cdot \pi }{3}$ is the key symmetry.


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