Prove that $2P$ is a flex of $C$ collinear with $P$ and $3P$ and $4P$ is a flex of $C$ collinear with $5P$ and $3P$. I encountered this problem from Conics and Cubics by Bix. Please help me answer this. Let $C$ be a nonsingular, irreducible cubic with a flex $O$. Add points (commutative) of $C$ with respect to $O$ as base point. Let $P$ be a point on $C$ of order 6. Prove that $2P$ is a flex of $C$ collinear with $P$ and $3P$. Prove that $4P$ is a flex of $C$ collinear with $5P$ and $3P$. Thanks!

The group law is equivalent to the following. If $L$ is any line and if it meets $C$ at $A,B,C$ (counted with multiplicity), then $A+B+C=O$. So, one easily sees that $2P, P, 3P$ are collinear and similarly for the other 3-tuple. Only thing remaining is to check that $2P, 4P$ are flexes. For $2P$, let $L$ be the tangent line at $2P$. Then $L$ meets $C$ at least twice at $2P$ and say at a point $A$. Then we have $2\cdot 2P+A=O$ and since $6P=O$, we see that $A=2P$, proving that $2P$ is a flex. Similarly for $4P$.