Prove the equation: $x^3+y^3+z^3-3xyz-1=0$ describes a cylindrical surface. Now I know that an equation describes a cylindrical surface iff the LHS is a function of two first order polynomials in $x,y,z$, that is we have an equation of the form: $F(a_1x+b_1y+c_1z, a_2x+b_2y+c_2z)=0$.
I can't for the life of me factorise this(yes, i tried wolframalpha) in a meaningful way.
yeah, rotated coordinates $$ u = (x+y+z) / \sqrt 3 , $$ $$ v = (-x+y) / \sqrt 2, $$ $$ w = (-x-y+2z) / \sqrt 6 \; . $$
I get $$ u(v^2 + w^2) = \frac{2}{\sqrt {27}} $$ For fixed $u,$ the section is a circle, but the radius shrinks as $u$ gets large. So, it is a surface of revolution around the line $x=y=z,$ but not really a cylinder
Try to get a picture of $$ z = \frac{1}{x^2 + y^2} $$