Pascal's theorem has a short proof using the Cayley–Bacharach theorem that given any 8 points in general position, there is a unique ninth point such that all cubics through the first 8 also pass through the ninth point. In particular, if 2 general cubics intersect in 8 points then any other cubic through the same 8 points meets the ninth point of intersection of the first two cubics. Pascal's theorem follows by taking the 8 points as the 6 points on the hexagon and two of the points (say, M and N in the figure) on the would-be Pascal line, and the ninth point as the third point (P in the figure).