Invertibility of the matrix whose elements are the cube of the distance of the indices. I would like to prove, for any integer $n&gt;1$, the invertibility of the $n\times n$ matrix $A$ whose elements are given by
$A_{ij}=|i-j|^3$, where $i$ and $j$ are the indices. 

To be clearer, for instance if $n=5$ I'm referring to the matrix
$$A=\begin{bmatrix}&amp;0&amp;1&amp;8&amp;27&amp;64\\&amp;1&amp;0&amp;1&amp;8&amp;27\\&amp;8&amp;1&amp;0&amp;1&amp;8\\&amp;27&amp;8&amp;1&amp;0&amp;1\\&amp;64&amp;27&amp;8&amp;1&amp;0\end{bmatrix}\,.$$

Such matrices should indeed be invertible (or at least it seems so looking at the determinant with Mathematica) for any $n&gt;1$ and I do believe that there's should be some easy way to show it. I had some look on invertibility for Toeplitz or Hankel matrices, but I couldn't find any helps there for now.
Actually I guess I've found a way to tackle the problem. Defining $q$ the matrix with elements $q_{ij}=|i-j|$, which is not hard to prove to be invertible, one has
$$q^{-1}\,A\,q^{-1}=M\,,$$
where $M$ is a sparse matrix (nearly tridiagonal), which can be easily shown to be invertible.
Now I'm looking for a smart way to prove that $M$ has such a simple structure. It's not hard to do it by direct calculations but it's a bit tedious and I'm sure there must be some smarter way to do it.
You can think of rank of matrix. 

If [mat A ] is to be invertible A has to to have full rank.