# Symetry in solutions of cubic and quartic equations

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 $(x^2+r_1)\cdot(x+r_2)=0; \{r_1,r_2\} \in R$
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.