0

Prove that the equation $x^3+y^3+z^3 -3xyz -1=0$ defines a surface by revolution and determine the equation for the axis of revolution.

I know I have to end up with something that looks like this: $(x-a)^2+(y-b)^2+(z-c)^2=r^2$

$dx+ex+fx = g$ but I don't know where to start...

(I am aware that there is a similar question on this site but I need a simpler answer than the one given there.)

caverac
  • 19,345
GiaFil7
  • 223
  • This is the one I'm refering to. I just need a simpler answer than the one that is given there... – GiaFil7 Jan 13 '18 at 17:22
  • The equation you've written down is the equation of a sphere, not of a surface of revolution. It's hard to write something too much simpler, other than by just pulling something out of thin air. You need a way of finding the axis of symmetry; that's what is going on in that answer. – Ted Shifrin Jan 13 '18 at 17:31
  • @TedShifrin I answered the same thing a couple of weeks ago. Hard to imagine how to do this without changing coordinates https://math.stackexchange.com/questions/2588202/prove-that-its-a-cylindrical-surfacemore-of-a-factorisation-question/2588252#2588252 – Will Jagy Jan 13 '18 at 18:19
  • I guess the answer by John Hughes at an earlier question, with some deleted comments it seems, amounts to: with constant $x+y+z = C,$ we get another constant $x^2 + y^2 + z^2 = B,$ and the intersection of the plane and sphere is a circle, a particular circle with a particular center... – Will Jagy Jan 13 '18 at 18:43

0 Answers0