Newton fractal roots in 3D space or arbitary dimension

[misterstarshine]

I've been having this idea for some years now but my math skill probably isn't good enough to make this idea come true.

I know that for the equation z^n - 1 = 0 has roots on the complex plane regardless of the exponent integer. I was thinking one could have an alternative formula that is different and would have roots on a sphere instead of just the unit circle in the complex plane. This would not be a part of the z^n - 1 = 0 family of formulas.

Take the newton basins fractal for the z^4 - 1 = 0 for example. My goal is to have a corresponding fractal with six roots on the unit sphere where the z^4 - 1 = 0 fractal is the cross sections in the xy-plane as well as in the xz-plane and in the yz-plane. I want the fractal to be fully symmetric in all directions so that any rotation by 90 degrees angle in space will generate the same figure. The z^4 - 1 = 0 has this symmetry in its own context in the complex plane. A rotation of 90 degrees in the complex plane generates the same fractal.

My own experiments so far have been with permutations of the formula for z^4 - 1 = 0 and trying to expand it to 3 dimensions. Here is one example by converting the fomula to two real equations where z = (x+iy)

  x^4 + y^4 -6*x^2*y^2 - 1 = 0

  4*x^3*y - 4*x*y^3 = 0

Now one could guess that the formula for the 3D counterpart with 6 roots on the unit sphere would be something like

  x^4 + y^4 +z^4 -6*x^2*y^2 --6*x^2*z^2 -6*y^2*z^2 - 1 = 0

  4*x^3*y - 4*x*y^3 = 0

  4*x^3*z - 4*x*z^3 = 0

The important thing here is that the z^4 - 1 = 0 figure should appear when setting any of the coordinates to 0 such as (x,y,0) or (x,0,z) or (0,y,z) when rendering. Even though I have succeded with this goal the symmetry for other cross sections fail and the 3D figure is not fully symmetric when rotating it 90 degrees around any of the unit axes. Any ideas how to obtain this goal if it even can be done?

Hope you find this idea intresting.