= Solution
On the nonsingular cubic, $x^3+y^3+z^3=-3\lambda xyz$. Substituting this into the Hessian equation leaves $-4(\lambda^3+1)xyz=0$. Since $\lambda^3+1\ne0$, a flex has a zero coordinate. The remaining cubic equation then gives the nine points
$$
\boxed{[0:1:-\omega^k],\quad[1:0:-\omega^k],\quad[1:-\omega^k:0],\qquad k=0,1,2,}
$$
where $\omega=e^{2\pi i/3}$. All are distinct, smooth, and lie on both the cubic and its <Hessian curve>, so they are exactly the <inflection points>. In particular the nine flexes are independent of $\lambda$ within the nonsingular family. This is the <flex configuration of the Hesse pencil>.
Back to article page
