The book Elliptic Curves Number Theory and Cryptography by Lawrence Washington contains this two-parts exercise:
a) Show that the constant term of a monic cubic polynomial is the negative of the product of the roots.
b) Use (a) to derive the formula for the sum of two distinct points $P1$, $P2$ in the case that the x-coordinates $x_1$ and $x_2$ are nonzero, as in Section 2.2. Note that when one of these coordinates is 0, you need to divide by zero to obtain the usual formula.
This question is an exact duplicate of this one that got no satisfactory answer. Considering that question is lacking the book context and that reviving a 3 year old question is very hard, I'm writing my own. Eventually, if this question gets an answer, we can make the old question a duplicate of this.
Part (a) is very easy but, shamefully, part (b) is giving me trouble.
In Section 2.2 the author is given two points $P1 = (x_1, y_1)$, $P2 = (x_2, y_2)$ (with $x_1 \neq x_2$) on an elliptic curve $y^2 = x^3 + Ax + B$.
They find the line through these points to be $y = m(x - x_1) + y_1$, with $m = (y_2 - y_1)/(x_2 - x_1)$.
They proceed with intersecting the line and the elliptic curve to obtain $x^3 - m^2x^2 + (A + 2m^2x_1 + 2my_1)x + (B - m^2x_1^2 + 2mx_1y_1 - y_1^2) = 0$.
Two of the roots of this equation are $x_1$ and $x_2$ by construction and the third is found by exploiting the fact that (for a monic cubic equation like this one) $x_1 + x_2 + x_3 = m^2$ and hence:
$$
x_3 = m^2 - x_1 - x_2\tag{1}\label{eq1}
$$
I believe part (b) is asking me to redo the passages above but using the fact that the roots $r_{1..3}$of a monic cubic equation $x^3 + ax^2 + bx + c = 0$ satisfy $r_1 r_2 r_3 = -c$.
I set: $$ \begin{equation} x_1 x_2 x_3 = m^2x_1^2 + y_1^2 - 2mx_1y_1 - B\tag{*}\label{eq0} \end{equation} $$
Then I substituted $y_1^2$ with $x_1^3 + Ax_1 + B$: $$\begin{equation}\begin{aligned} x_1 x_2 x_3 &= m^2x_1^2 + x_1^3 + Ax_1 + B - 2mx_1y_1 - B \\ &= x_1(m^2x_1 + x_1^2 + A - 2my_1) \\ x_2 x_3 &= m^2x_1 + x_1^2 + A - 2my_1 \\ x_3 &= \frac{m^2x_1 + x_1^2 + A - 2my_1}{x_2} \\ \end{aligned}\end{equation}\tag{2}\label{eq2}$$
But now I'm stuck.
Equation $\eqref{eq2}$ seems to be correct as it gives me the same values of $\eqref{eq1}$ (tested with this code) but I don't see how to turn $\eqref{eq2}$ into $\eqref{eq1}$.
I also tried factoring the binomial square in $\eqref{eq0}$, replacing $B$ with $y_2^2 - x_2^3 - Ax_2$ and rewriting $m$ but got nowhere.
How to get $\eqref{eq1}$ from $\eqref{eq2}$?