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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.

enter image description here

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?

Blue
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InsideOut
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    Graphing software typically plots implicit relations by testing points to see what works. As WillJagy's answer notes, the polynomial $x^3+y^3+z^3-3xyz=0$ decomposes as the plane equation $x+y+z=0$ and effectively the line equation $x=y=z$. The graphing software seems to have done a good job of detecting the plane; perhaps the engine actually did the factorization and knew how to interpret the plane. It's not surprising that the test points miss the one-dimensional line, and/or that the software didn't "see" the other factor as representing a line. – Blue May 03 '18 at 18:56

1 Answers1

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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 $x = y = z$

Will Jagy
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    Your original real valued factorization was immensely helpful in understanding the situation. – Cedron Dawg May 03 '18 at 18:51
  • @CedronDawg take a look at this one: https://math.stackexchange.com/questions/2678486/how-to-show-that-if-x-y-z-are-rational-numbers-satisfying-x-y-z3-9/2678675#2678675 I was also looking for a visual... – Will Jagy May 03 '18 at 18:56
  • That's interesting too. Strange that it got marked "off-topic" with 6 up votes going for it. – Cedron Dawg May 03 '18 at 19:38