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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.

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

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math.stackexchange.com is a better place for these questions. –  Mariano Suárez-Alvarez Feb 21 '13 at 8:51
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2 Answers 2

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

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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 $(1,1,1)$.
           Implicit Surface

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