Is this surface diffeomorphic to a sphere(S^2)? [closed]

let $f:R^3 -> R \ \ \ \ \ \ \ \ f(x,y,z)=x^4 + y^6 +z^8 \\$

$M = f^{-1}(1)$

Is M is diffeomorphic to a sphere $S^2$ ?

I tried to solve this problem, but I realized that I have no tools to solve it.

Constant rank theorem tells me M is a smooth 2 dimensional manifold, but does not tell me how it looks like.

And more generally, when $N = \{ x,y,z \in R^3 | ax^n + by^m + cz^l = 1\}$ is diffeomorphic to a sphere? What tools can I use to solve this problem?

Thank you for reading. Hoping get some shedding light in your reply.

-

closed as off topic by Chris Gerig, Benoît Kloeckner, Mariano Suárez-Alvarez♦, Martin Brandenburg, Willie WongFeb 21 '13 at 12:09

Questions on MathOverflow are expected to relate to research level mathematics within the scope defined by the community. Consider editing the question or leaving comments for improvement if you believe the question can be reworded to fit within the scope. Read more about reopening questions here.If this question can be reworded to fit the rules in the help center, please edit the question.

math.stackexchange.com is a better place for these questions. – Mariano Suárez-Alvarez Feb 21 '13 at 8:51

2 Answers

It is. The homeomorphism from $X$ to the unit sphere is $$x\mapsto \mathrm{sign}(x)\cdot x^2,\ y\mapsto y^3,\ z\mapsto\mathrm{sign}(z)\cdot z^4.$$

-

Here is a plot of the surface $x^4+y^6+z^8 = 1$. It is approximately a cube with opposite corners $(-1,-1,-1)$ and $(1,1,1)$.

-