I have found this relevant question here, which gives the result, but doesn't show how to get to the neatly factored answer, which completes the proof.
A parabola goes through the points $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ such that $x_1 \neq x_2 \neq x_3 $. To show that there is exactly one second degree polynomial solution, we can plug the points into the generic form $y = ax^2+bx+c$ and put those in matrix equation form of Ax = b, where $$ A =\begin{bmatrix} x_1^2 & x_1 & 1 \\ x_2^2 & x_2 & 1 \\ x_3^2 & x_3 & 1 \\ \end{bmatrix} $$
If we can show that $det(A) \neq 0$, this is equivalent to showing that there's exactly one solution to this system.
Using column expansion, $$det(A) = (x_2^2 \cdot x_3 - x_2 \cdot x_3^2) - (x_1^2 \cdot x_3 - x_1 \cdot x_3^2) + (x_1^2 \cdot x_2 - x_1 \cdot x_2^2)$$
This can be factored to $$(x_2 \cdot x_3)(x_2 - x_3) - (x_1 \cdot x_3)(x_1-x_3)+(x_1 \cdot x_2)(x_1 - x_2)$$
I can continue with $$x_3[x_2(x_2-x_3)-x_1(x_1-x_3)] + x_1[x_2(x_1-x_2)]$$
How do you continue factoring to get to the form $$(x_1 - x_2)(x_1-x_3)(x_2-x_3)$$?