When do 10 points lie on a cubic? (a la Pascal) Pascal theorem gives a necessary and sufficient condition for 6 points to lie on a quadratic curve in terms of lines: $A_1,\ldots,A_6$ lie on a quadric iff 3 intersection points $A_1A_2\cap A_4A_5$, $A_2A_3\cap A_5A_6$, $A_3A_4\cap A_6A_1$ are collinear.
Is there a theorem of this kind for cubic curves: a necessary and sufficient condition for 10 points of a plane to lie on a cubic curve, formulated in terms of conics and lines?
 A: (After Jan-Magnus Økland's comment — thanks again! — I've been able to find an answer.)
Theorem ([1], thm CB4). Let $X_1$, $X_2$ be plane curves of degrees $d$ and $e$ respectively, meeting in a collection of $d\cdot e$ distinct points $\Gamma$. If $C$ is any plane curve of degree $d+e-3$ containing all but one point of $\Gamma$, then $C$ contains all of $\Gamma$.
For $d=e=3$ ($d+e-3=3$) this is Chasles's theorem, which implies Pascal theorem.
And $d=e=4$ ($d+e-3=5$) implies the following recipe.
Theorem (Traves, Welhau; [2]). Split 10 points into two groups of 5, giving red and blue degree 4 curves (consisting of two conics each) and find intersections of red and blue curves. Our 10 points lie on a cubic iff 6 auxiliary points lie on a conic.

// Caveat. If you'll try to use this recipe in, say, Geogebra you'll see that auxiliary conics frequently don't have enough real intersection points.  


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*[1]: D. Eisenbud, M. Green and J. Harris. Cayley–Bacharach theorems and conjectures// Bull. Amer. Math. Soc. 33 (1996), 295–324

*[2]: W. Traves. Ten Points on a Cubic (presentation).

