My question arises from another question here. The OP asked to computed the tangent plane to the surface $S$ given by the implicit equation
$$x^3+y^3+z^3-3xyz=0$$
at $(1,1,1)$. In order to get an idea of how $S$ should be, I tried to plot it by obtaining the image in the figure below.
What puzzles me is that: the graph of $S$ does not pass through the point $(1,1,1)$ (the blue point); but yet $(1,1,1)$ is obviously a solution of $x^3+y^3+z^3-3xyz=0$. I immediately thought there was a mistake then I checked the equation more than one time. Hence I thought that my program makes a fallacious plot, but different programs provide the same plot. Thus: how is it possible?
The polynomial factors. One factor is $x+y+z.$ the other factor is $$ \frac{1}{2} \left( (x-y)^2 + (y-z)^2 + (z-x)^2 \right) $$ which gives the entire line $x=y=z$
Ummm. A homogeneous cubic in three variables factors completely (over the complexes) if the determinant of the Hessian matrix, which is itself a cubic form, is a constant multiple of the original. Taking $$ \omega = \frac{-1 + i \sqrt 3}{2} $$ so that $\omega^3 = 1,$ the complete factorization is $$ (x+y+z)(x+ y \omega + z \omega^2)(x + y \omega^2 + z \omega) $$
The Hessian is $$ \left( \begin{array}{ccc} 6x&-3z&-3y \\ -3z&6y&-3x \\ -3y&-3x&6z \\ \end{array} \right) $$ with determinant $$ -54 \left( x^3 + y^3 + z^3 - 3 x y z \right) $$
This example is well known on this website. The most difficult example I have done had all real but irrational coefficients, How to show that if $x, y, z$ are rational numbers satisfying $(x + y + z)^3 = 9(x^2y + y^2z +z^2x)$, then $x = y = z$