Let six points $A, A', B, B', C, C'$ lie on a conic and a cubic. Let a conic through $B, B', C, C'$ and meets the cubic again at $A\_1, A\_2$. Let a conic through $C, C', A, A'$ and meets the cubic again at $B\_1, B\_2$. Let a conic through $A, A', B, B'$ and meets the cubic again at $C\_1, C\_2$. Then six points $A\_1,A\_2, B\_1, B\_2, C\_1, C\_2$ lie on a conic. In this case if we let $A\_1, B\_1, C\_1$ are collinear then $A\_2, B\_2, C\_2$ are collinear.


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**Special case 1, The cubis are three lines.** Let six points $A, A', B, B', C, C'$ lie on a conic. 
Let a conic through $B, B', C, C'$ and meets line $AA'$ at $A\_1, A\_2$
Let a conic through $C, C', A, A'$ and meets line $BB'$ at $B\_1, B\_2$
Let a conic through $A, A', B, B'$ and meets line $CC'$ at $C\_1, C\_2$.
Then six points $A\_1,A\_2, B\_1, B\_2, C\_1, C\_2$ lie on a conic. In this case if $A\_1, B\_1, C\_1$ are collinear then $A\_2, B\_2, C\_2$ are collinear.


[Check the special case 1 by ggb](http://tube.geogebra.org/material/simple/id/1563641)


**Special case 2, A conic and a cubic:** Let a conic meets a cubic at six points $A, B, C, A', B', C'$. Let $AB', A'B$ cut the cubic at $C\_1, C\_2$ respectively; Let $BC', B'C$ cut the cubic again at $A\_1, A\_2$; Let $AC', A'C$ cut the cubic at $B\_2, B\_1$ then $A\_1, B\_1, C\_1$ are collinear and $A\_2, B\_2, C\_2$ are collinear.


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**Special case 3, The conics are lines:** Let two lines meet a cubic at $A, B, C$; and $A', B', C'$ respectively. Let $AB', A'B$ cut the cubic at $C\_1, C\_2$ respectively; Let $BC', B'C$ cut the cubic again at $A\_1, A\_2$; Let $AC', A'C$ cut the cubic at $B\_2, B\_1$ then $A\_1, B\_1, C\_1$ are collinear and $A\_2, B\_2, C\_2$ are collinear.


[Check the special case 3 by ggb](http://tube.geogebra.org/m/1566001)


**Special case 4, The cubic are a line and a conic:** Let the cubic is a conic ($\gamma$) through four points $A, A', B, B'$ and line $CC'$ on the plane). Let points $A, B, C, A', B', C'$ lie on another conic. Let $AB', A'B$ meet $CC'$ at $C\_1, C\_2$; let $AC', A'C, BC', B'C$ meet conic $\gamma$ at $B\_2, B\_1, A\_1, A\_2$. Then six points $A\_1, A\_2, B\_1, B\_2, C\_1, C\_2$ lie on a conic.


[Check the special case 4 by ggb](http://tube.geogebra.org/m/1566085)



A conjecture like Cayley–Bacharach theorem
If the cubic is smooth and given the usual group law with an inflexion as identity, the condition that given six points $P\_1,\ldots,P\_6$ on the cubic they also lie on a conic is $P\_1+\cdots+P\_6=0$. In your case 
$$A+A'+B+B'+C+C'=0$$
$$A\_1+A\_2+B+B'+C+C'=0$$
$$A+A'+B\_1+B\_2+C+C'=0$$
$$A+A'+B+B'+C\_1+C\_2=0$$
Adding the last three equations and using the first, gives $A\_1+A\_2+B\_1+B\_2+C\_1+C\_2=0$, which is what you want. The general case probably follows by degenerating the cubic.


