Solving cubic equation - deriving the resolvent explicitly

Question

I'm trying to fill in the details of the systematic approach of deriving the quadratic resolvent for the cubic equation according to this part of the 3Blue1Brown talk: https://youtu.be/aaW30_f2on0?t=4374. To sum up, given a cubic

$$x^3 + px + q$$

the idea is to find some polynomial expression in its roots $a, b, c$ with the orbit of size 2 under the actions of $S_3$. The suggested expressions

$$A = a^2b + b^2c + c^2a$$ $$B = a^2c + b^2a + c^2b$$

indeed constitute such an orbit. Therefore the quadratic polynomial $$(y - A)(y - B) = y^2 - (A +B)y + AB$$ has coefficients in the $p, q$. These coefficients can be expressed in the elementary symmetric polynomials ($e_1=0, e_2=p, e_3=-q$) as follows (I used Sage for the reduction): $$A+B = -3e_3 = 3q$$ $$AB = e_2^3 + 9e_3^2 = p^3 + 9q^2$$ Computing the discriminant of this quadratic equation yields: $$D = -4p^3 - 27q^2$$ This looks promising since this is up to a constant factor (and sign) the same expression as the one appearing in the final cubic formulas. Moreover, it coincides with the official discriminant of the cubic equation.

The question is, however, how to proceed from this point, i.e. how to derive the expressions for the roots $a, b, c$ of the original cubic. The only relations between the roots we have so far are the original expressions

$$A = a^2b + b^2c + c^2a$$ $$B = a^2c + b^2a + c^2b$$

We can now express the left-hand sides as (radical) expressions of $p$ and $q$ but of course this does not fully determine $a, b, c$.

Answer 1

Following on Jyrki Lathonen's hint, render the polynomial power expansion below. The quantity $\omega$ is taken to be the unit root $(-1+i\sqrt3)/2$, thus $\omega^3=1$.

$(a+\omega b+\omega^2c)^3=a^3+b^3+c^3+3[a^2(\omega b)+a^2(\omega^2c)+a(\omega b)^2+a(\omega^2c)^2+(\omega b)^2(\omega^2c)+(\omega b)(\omega^2c)^2]+6a(\omega b)(\omega^2c)$

$=(a^3+b^3+c^3)+3\omega(a^2b+b^2c+c^2a)+3\omega^2(ab^2+bc^2+ca^2)+6abc$

$=(a^3+b^3+c^3)+3\omega A+3\omega^2 B-6q$

where we have plugged in $abc=-q$.

Now comes the subtle part. We have the factorization

$a^3+b^3+c^3-3abc=\color{blue}{(a+b+c)}(a^2+b^2+c^2-ab-ac-bc)$

and $a+b+c=0$! So $a^3+b^3+c^3=3abc=-3q$ and we finally have

$(a+\omega b+\omega^2c)^3=3\omega A+3\omega^2 B-9q$

We repeat the above with $\omega^2$ in place of $\omega$, so that $(\omega^2)^2=\omega^4=\omega$; thus

$(a+\omega^2b+\omega c)^3=3\omega^2 A+3\omega B-9q.$

There is, actually, a better way to render $(a+\omega^2b+\omega c)$. Form the product

$(a+\omega b+\omega^2c)(a+\omega^2b+\omega c)=(a^2+b^2+c^2)-(ab+ac+bc).$

Squaring the relation $a+b+c=0$ for the depressed cubic leads to $a^2+b^2+c^2=-2(ab+ac+bc)$, and we also have the symmetric sum $(ab+ac+bc)=p$. Thus

$(a+\omega b+\omega^2c)(a+\omega^2b+\omega c)=-3p.$

This avoids introducing a second cube root, and allows the roots $a,b,c$ to be rendered in terms of branches of a single cube root.

So

$a+b+c=0$

$a+\omega b+\omega^2c=\sqrt[3]{3\omega A+3\omega^2 B-9q}$

$a+\omega^2b+\omega c=-3p/\sqrt[3]{3\omega A+3\omega^2 B-9q}$

We now have a linear system with a coefficient matrix having a nonzero determinant (we compute the determinant as $3i\sqrt3$), so we have a unique solution for $a,b,c$. Dfferen branches of the single cube root $\sqrt[3]{3\omega A+3\omega^2 B-9q}$ correspond to different orderings of the same roots $,b,c$. When $A$and $B$ are real and distinct, the solution will contain complex cube roots, whose real and imaginary parts are not accessible by algebraic methods.

Answer 2

One standard approach to solving the cubic is to write $x=u+v$, in which case we have a solution provided $uv=-p/3$ and $u^3+v^3=-q$. Then $u^3,v^3$ are the solutions of the auxiliary quadratic $t^2+qt-p^3/27$, and one root of the original cubic is of course $\sqrt[3]{u^3}+\sqrt[3]{v^3}$.

Comparing this quadratic to yours, we have $A=\lambda u^3+\frac12q(3+\lambda)$ and $B=\lambda v^3+\frac12q(3+\lambda)$, where $\lambda^2=-27$. It is then clear that the choices for $A$ and $B$ are not necessarily the most appropriate ones for obtaining the roots for the cubic.

Note also that one often takes $\delta=(a-b)(a-c)(b-c)$, so that $\delta$ is fixed by all even permutations of $a,b,c$, and sent to its negative under odd permutations. Expanding out we have $\delta=B-A=\lambda(v^3-u^3)$.