Third point of intersection is also a point of inflection? Let $C \subset \mathbb{P}_2$ be a nonsingular cubic. If $L$ is a line through two distinct points of inflection on $C$, how do I show that the third point of intersection is also a point of inflection?

This is Problem 2 in section 5.7, "Elliptic Curves," of I. Niven, H. S. Zuckerman, H. L. Montgomery, An Introduction to the Theory of Numbers, 5th ed., Wiley (New York), 1991.  I use the nomenclature from that book.
After I worked out this answer, I realized that it is the same as Lubin's.
$C$ is elliptic.  Let $\mathbf{A}$ and $\mathbf{B}$ be the two distinct inflection points and $\mathbf{AB}$ the third point.  Choose $\mathbf{0}$ to be an inflection point.  We use the following facts.  The proof of the main result is farther down.


*

*Fact 1: $\mathbf{AA} = \mathbf{A}$, $\mathbf{BB} = \mathbf{B}$, and $\mathbf{00} = \mathbf{0}$.

*Converse of Fact 1:  If $\mathbf{PP} = \mathbf{P}$, then $\mathbf{P}$ is an inflection point.

*Fact 2:  $\mathbf{P}(\mathbf{0P}) = \mathbf{0}$.  Indeed, $\mathbf{0P}$ is the third point on the line through $\mathbf{0}$ and $\mathbf{P}$.  The line through $\mathbf{P}$ and $\mathbf{0P}$ is the same line, so the third point on the line is the original point $\mathbf{0}$.

*Fact 3: $\mathbf{A} + \mathbf{A} + \mathbf{A} = \mathbf{0}$, proved as follows:
\begin{align}
\mathbf{A} + \mathbf{A} + \mathbf{A} & = \mathbf{A} + \mathbf{0(AA)} &&\text{definition of +}\\
& = \mathbf{A} + \mathbf{0(A)} &&\text{Fact 1}\\
& = \mathbf{0(A(0A))} &&\text{definition of +}\\
& = \mathbf{00} &&\text{Fact 2}\\
& = \mathbf{0} &&\text{Fact 1}
\end{align}
Likewise, $\mathbf{B} + \mathbf{B} + \mathbf{B} = \mathbf{0}$.

*Converse of Fact 3:  If $\mathbf{P} + \mathbf{P} + \mathbf{P} = \mathbf{0}$, then $\mathbf{P}$ is an inflection point.  To prove that, we  have
\begin{align}
\mathbf{P} + \mathbf{P} + \mathbf{P} & = \mathbf{0}\\
\mathbf{0(P(0(PP)))} & = \mathbf{00} &&\text{definition of + on left; Fact 1 on right}\\
\mathbf{P(0(PP))} & = \mathbf{0} &&\text{by uniqueness}
\end{align}
We also know that $\mathbf{P(0P)} =  \mathbf{0}$ by Fact 2.  Comparing these last two equations, we see that $\mathbf{PP} = \mathbf{P}$ by uniqueness.  Hence, $\mathbf{P}$ is an inflection point by the converse of Fact 1.

*Fact 4: $\mathbf{A} + \mathbf{B} + \mathbf{AB} = \mathbf{0}$ is proved at The sum of three colinear rational points is equal to $O$.  (That proof applies to general points, not necessarily rational.)
We are now ready for the proof of the main result:
\begin{align}
\mathbf{0} & = \mathbf{A} + \mathbf{A} + \mathbf{A} + \mathbf{B} + \mathbf{B} + \mathbf{B} &&\text{Fact 3}\\
& = \mathbf{A} + \mathbf{B} + \mathbf{A} + \mathbf{B} + \mathbf{A} + \mathbf{B} &&\text{+ is commutative}\\
& = -\mathbf{AB} - \mathbf{AB} - \mathbf{AB} &&\text{Fact 4}
\end{align}
Because $-\mathbf{0} = \mathbf{0}$, we have that $\mathbf{AB} + \mathbf{AB} + \mathbf{AB} = \mathbf{0}$.  Thus, by the converse of Fact 3, $\mathbf{AB}$ is an inflection point.
