Prove that $x^3+y^3+z^3-3xyz=1$ defines a surface of revolution

Question

Prove that the equation $x^3+y^3+z^3-3xyz=1$ defines a surface of revolution and find the analytical equation of its axis of revolution.

I think that I need to apply Euler's formula, so that I get rid of the third-grade polynomial there: $x^3+y^3+z^3-3xyz=1 \Leftrightarrow (x+y+z)(x^2+y^2+z^2-xy-xz-yz)=1$ but then I stuck on how to prove it defines a surface of revolution.

The textbook notes that: $f(x,y,z)=0$ defines a surface of revolution around the axis with equation $\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}$ if and only if it can be written as a polynomial of $(x-x_0)^2+(y-y_0)^2+(z-z_0)^2$ and $ax+by+cz$. Any hint for that one?

Answer 1

$x^2+y^2+z^2−xy−xz−yz = \mathbf x^T A \mathbf x$

Diagonlize $A$ and find and ortho-normal basis.

As it turns out $(1,1,1)^T$ is an eigenvector of A. The other eigenvalue is a duplicated eigenvalue. And between those two bits of information, that suggests a surface of revolution.

Along similar lines:

$u = \frac 1{\sqrt 3} (x + y + z)\\ v = \frac 1{\sqrt 2} (x+y)\\ w = \frac 1{\sqrt 6} (x+y+2z)$

make this substitution and $(x+y+z)(x^2+y^2 + z^2 - xy-xz - yz) = 1$ becomes

$\frac {3\sqrt 3}{2} u (v^2 + w^2) = 1$

$v = r \cos \theta\\ w = r \sin \theta$

$u= \frac 2{3r\sqrt3}$

Answer 2

Hint: The last three terms -- $xy + xz + yz$-- can be simplified by looking at $(x+y+z)^2$.