 Flex configuration of the Hesse pencil

For ${\omega }^3=1$ primitive and $k=0,1,2$, these nine points are the inflection points of every nonsingular finite member of the Hesse pencil. On the cubic, substituting the equation into its Hessian curve gives $-4({\lambda }^3+1)xyz=0$. Nonsingularity forces one coordinate to vanish, and the remaining two cubes sum to zero, producing exactly the nine displayed points. They are the fixed base configuration of the pencil, independently of its smooth parameter.