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If planes are described as: $\mathbf{n} \cdot (\mathbf{r}-\mathbf{r_0})=0$

And quadric surfaces can be described as: $\mathbf{x}^T\mathbf{A}\mathbf{x} = 0$ (with $\mathbf{x} = \begin{bmatrix} x && y && z && 1\end{bmatrix}^T$ and $\mathbf{A}$ is a symmetric matrix)

Then what is the name and compact form for a surface in the form of $Ax^3 + By^3 + Cz^3 + Dx^2z + \ldots = 0$

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Cubic hypersurface. –  user27126 May 4 '13 at 23:40
    
@Sanchez, is cubic *hyper*surface still applicable when the polynomial is still an implicit function of three variables? –  Damien May 8 '13 at 22:45
    
Oh, I forgot to add - @Sanchez, if you answer the above comment, and provide a compact form for the equations, add it an an answer and you'll have yourself a nice easy accepted solution. –  Damien May 8 '13 at 22:48

1 Answer 1

up vote 3 down vote accepted

The name is cubic hypersurface. (Hypersurface generally means something of one lower dimension, so it's not relevant whether your equation is implicit. I would also call your quadratic surface as quadratic hypersurface and plane as hyperplane once we move to higher dimensions. It's okay for you to remove "hyper" though.)

I don't think there is a more compact form for the equation, although naively you can do a change of variable to cancel out a few coefficients.

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Now that I know what to look for Algebraic Surfaces is a good place to start of surfaces of other orders. Also, Cubic Surface yields further results that are not Cubic hyper surfaces. –  Damien May 9 '13 at 22:20

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