Homographic relation between two roots of a cubic

Question

Consider a cubic equation $ x^3 + 3a x^2 + 3 b x + c=0$ with distinct roots. Show that any two roots $x$, $y$ are connected by a homographic relation

$$(a^2-b) x y + \frac{1}{2}\ (\ (a b-c+\delta) x + ( a b -c-\delta) y\ ) + (b^2 - a c)=0 \ \ \ \ \ \ (*)$$

where $\delta = \pm\frac{\sqrt{\Delta}}{9}$ and $\Delta$ is the discriminant of the cubic.

Notes:

1. This appears in Burnside and Panton -- The Theory of Equations, with the same title ( the notations changed a bit)

2. Example: for the equation $x^3 - 4 x + x + 1=0$ with discriminant $13^2$ we get $ y= \frac{1}{1-x}$ or $y = \frac{x-1}{x}$

3. For the case of a cubic equation

$$(x+w)^3 - \frac{3(u^3+v^3)}{u+v} (x+w) + (u^3+v^3)$$

with discriminant $\Delta = 9^2 (u-v)^2(u^2-u v + v^2)^2$ we get the transformations of the Galois group $\phi_{0,1,2}(x)$, with $\phi_0(x) = x$, \begin{eqnarray} \phi_{1}(x) = (u-w) + \frac{((v-w) - (u-w))^2}{v-w-x}\\ \phi_{2}(x) = (v-w) + \frac{((u-w) - (v-w))^2}{u-w-x} \end{eqnarray}

Note that $\phi_i \circ \phi_j = \phi_{i+j}$, where $i,j \in \mathbb{Z}/3$.

The case $w=0$ is even nicer.

4. Consider a matrix $M_{p,q}= \left(\begin{matrix} -\frac{1}{2}-p & \frac{3/4 + p^2}{q}\\ q & -\frac{1}{2}+p\end{matrix} \right)$ with characteristic polynomial $t^2 + t + 1$. We have $M_{-p,-q} =M_{p,q}^2 = M_{p,q}^{-1}$. If we specialize $p = \frac{a b - c}{2 \delta}$, $q=\frac{a^2-b}{\delta}$ we get the transformations of the Galois group of the cubic $x^3 + 3 a x^2 + 3 b x + c$. These transformations form a cyclic group of order $3$ no matter what $\delta$ is. However, for the mappings obtained from the homographic relation $(*)$ we need the specific condition on $\delta$ to get order $3$.

5. There are several question on this site (here and here) about finding the roots of a cubic with square discriminant in terms of one root. The expressions in polynomial form seems more involved. Moreover, every element of a cubic extension $K \subset K(x)$ is a fraction $\frac{\alpha x + \beta}{\gamma x + \delta}$, hence the problem finding such expressions for the other roots.

6. This seems to be on the beaten path, so is posted as reference.

Any feedback would be appreciated!

$\bf{Added:}$ Let us note that for the equation

$(x+w)^3 - \frac{3(u^3 + v^3)}{u+v}(x+w) + (u^3 + v^3)$ the homographic transforms are given by the matrix

\begin{eqnarray} \left (\begin{matrix} \frac{w-u}{u-v} & \frac{ (u-v)^2+(v-w)^2+(w-u)^2}{2(u-v)}\\ -\frac{1}{u-v} & \frac{v-w}{u-v} \end{matrix} \right) \end{eqnarray}

Let us write $\alpha = u-v$, $\beta= v-w$, $\gamma = w-u$, with $\alpha + \beta + \gamma = 0$. So we have a matrix

\begin{eqnarray} \left (\begin{matrix} \frac{\gamma}{\alpha} & \frac{\alpha^2 + \beta^2 + \gamma^2}{2 \alpha} \\ -\frac{1}{\alpha} & \frac{\beta}{\alpha} \end{matrix} \right ) \end{eqnarray} with characteristic polynomial $t^2 + t+1$. With $\alpha$, $\beta$, $\gamma$ of sum $0$ we can parametrize all such matrices uniquely. So now we see a way to get all the cubics with Galois group containing a specific homographic transform of order $3$ ( maybe some problems here with lifting to SL2 here, I'll think about it later).

Answer 1

Multiplying by $18$ and rearranging, the equality to be proved is:

$$ 18(a^2-b) x y + 9(a b-c)(x+y) \pm 9\delta(x - y) + 18(b^2 - a c)=0 \tag{1} $$

Let the third root be $z$, then the elementary symmetric polynomials in $x,y$ can be written in terms of $z$ as:

$$ \begin{align} x+y = -3a-z \tag{2} \\ xy = -\frac{c}{z} \tag{3} \end{align} $$

Choosing the sign for $\pm \sqrt{\Delta}$ such that $\delta = \frac{1}{9}(x-y)(y-z)(z-x)\,$:

$$ \begin{align} -9\delta(x-y) &= (x-y)^2(z-x)(z-y) \\ &= \big((x+y)^2 - 4xy\big)(z^2 - (x+y)z + xy) \\ &= \big(z^2 + 6a z + 9a^2 + \frac{4c}{z}\big)\big(2z^2 + 3 a z - \frac{c}{z}\big) \\ &= \frac{1}{z^2}(z^3+6az^2+9a^2z+4c)(2z^3 + 3 a z^2 - c) \tag{4} \end{align} $$

$\big($Another way to read $(4)$ is as an expression for $\,|x-y|=\dfrac{\pm\,9\delta z}{2z^3 + 3 a z^2 - c}\,$ in terms of $\,z, \delta\,$.$\big)$

Substituting $(2),(3),(4)$ in $(1)\,$:

$$ - 18(a^2-b) \frac{c}{z} - 9(a b-c)(3a + z) - \frac{1}{z^2}(z^3+6az^2+9a^2z+4c)(2z^3+3az^2-c) + 18(b^2 - a c) = 0 $$

Clearing the denominators gives a sextic in $z$, which has the original cubic as a factor (verified by WA):

$$ \begin{align} &- 18(a^2-b) c z - 9(a b-c)(3a + z)z^2 - (z^3+6az^2+9a^2z+4c)(2z^3+3az^2-c) + 18(b^2 - a c)z^2 \\ &\;\;=\;\; -(z^3 + 3 a z^2 + 3 b z + c) \left(2 z^3 + 9 a z^2 + 3(3 a^2 - 2 b) z - 4 c\right) \\ &\;\;=\;\; 0 \end{align} $$