Hesse pencil ($x^3+y^3+z^3+3\lambda xyz=0$)

This pencil consists of projective cubic curves generated by $x^3+y^3+z^3$ and $xyz$. The displayed finite parameter gives a nonsingular cubic exactly when ${\lambda }^3\ne -1$; the parameter at infinity gives the triangle $xyz=0$. Multiplying the three gradient equations proves the finite singularity criterion. The Hessian curve of a member has equation ${\lambda }^2(x^3+y^3+z^3)-({\lambda }^3+4)xyz=0$, so the Hessian stays in the same 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.