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\begin{vmatrix}5- 3 \cdot 2^4 & 2^1 & 2^2 & 2^3 \\ 10 - 3 \cdot 3^4 & 3^1 & 3^2 & 3^3 \\ 2 -3 (-1)^4 & (-1)^1 &(-1)^2 & (-1)^3 \\ 37-3(-6)^4 & (-6)^1 & (-6)^2 & (-6)^3 \\ \end{vmatrix}
This is a determinant that came up while I was doing a problem.. it has almost the structure of the Vandermonde determinant but I can't see if there are easy simplifications. Any help/ hint is appreciated.
It is easy if you see $5=2^2+1$.
$\begin{vmatrix}5- 3 \cdot 2^4 & 2^1 & 2^2 & 2^3 \\ 10 - 3 \cdot 3^4 & 3^1 & 3^2 & 3^3 \\ 2 -3 (-1)^4 & (-1)^1 &(-1)^2 & (-1)^3 \\ 37-3(-6)^4 & (-6)^1 & (-6)^2 & (-6)^3 \\ \end{vmatrix} = $
$\begin{vmatrix} 2^4 & 2^1 & 2^2 & 2^3 \\ 3^4 & 3^1 & 3^2 & 3^3 \\ (-1)^4 & (-1)^1 &(-1)^2 & (-1)^3 \\ (-6)^4 & (-6)^1 & (-6)^2 & (-6)^3 \\ \end{vmatrix}\cdot (-3)+$
$\begin{vmatrix}2^2 & 2^1 & 2^2 & 2^3 \\ 3^2 & 3^1 & 3^2 & 3^3 \\ (-1)^2 & (-1)^1 &(-1)^2 & (-1)^3 \\ (-6)^2 & (-6)^1 & (-6)^2 & (-6)^3 \\ \end{vmatrix} +$
$\begin{vmatrix}1 & 2^1 & 2^2 & 2^3 \\ 1 & 3^1 & 3^2 & 3^3 \\ 1 & (-1)^1 &(-1)^2 & (-1)^3 \\ 1 & (-6)^1 & (-6)^2 & (-6)^3 \\ \end{vmatrix} $
Do an expansion down the first column, then you have subdeterminants to calculate that ARE all essentially Vandermonde.
What I mean:
$$\begin{vmatrix}5- 3 \cdot 2^4 & 2^1 & 2^2 & 2^3 \\ 10 - 3 \cdot 3^4 & 3^1 & 3^2 & 3^3 \\ 2 -3 (-1)^4 & (-1)^1 &(-1)^2 & (-1)^3 \\ 37-3(-6)^4 & (-6)^1 & (-6)^2 & (-6)^3 \\ \end{vmatrix} = (5-3\cdot2^{4}) \begin{vmatrix} 3^1 & 3^2 & 3^3 \\ (-1)^1 &(-1)^2 & (-1)^3 \\ (-6)^1 & (-6)^2 & (-6)^3 \\ \end{vmatrix} + \ldots \\ = (5-3\cdot2^{4})\cdot3\cdot(-1)\cdot(-6) \begin{vmatrix} 1 & 3 & 3^2 \\ 1 &(-1) & (-1)^2 \\ 1 & (-6) & (-6)^2 \\ \end{vmatrix} + \ldots$$
(Also, since this is a $4\times 4$ matrix it should be easy to compute the determinant without any special tricks).
