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I need to find the value of the following determinant.

$$\det\begin{pmatrix}x^2&(x+1)^2&(x+2)^2\\ \:y^2&(y+1)^2&(y+2)^2\\ \:z^2&(z+1)^2&(z+2)^2\end{pmatrix}$$

By long calculations (by minors and properties), I found that the value is

$$4x^2y-4x^2z-4xy^2+4xz^2-4yz^2+4y^2z=4(x-y)(x-z)(y-z)$$

I'm wondering if there is an easier way to calculate this. Any help is welcome.

Rodrigo de Azevedo
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Daniel G. Ulate Montero
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    Alt. hint: the determinant $\Delta$ is a quadratic in $x$, and it is $0$ for $x=y$ and $x=z$, so it is divisible by $(x-y)(x-z)$. By symmetry, it must be of the form $\Delta = C,(x-y)(x-z)(y-z)$, and all that's left to do is determine the constant $,C,$. – dxiv Jul 30 '21 at 06:39
  • Related but different. – Jyrki Lahtonen Jul 30 '21 at 07:13

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Here's a possibility:\begin{align}\begin{vmatrix}x^2&(x+1)^2&(x+2)^2\\y^2&(y+1)^2&(y+2)^2\\ z^2&(z+1)^2&(z+2)^2\end{vmatrix}&=\begin{vmatrix}x^2&x^2+2x+1&x^2+4x+4\\y^2&y^2+2y+1&y^2+4y+4\\ z^2&z^2+2z+1&z^2+4z+4\end{vmatrix}\\&=\begin{vmatrix}x^2&2x+1&4x+4\\y^2&2y+1&4y+4\\ z^2&2z+1&4z+4\end{vmatrix}\\&=2\begin{vmatrix}x^2&2x+1&2x+2\\y^2&2y+1&2y+2\\ z^2&2z+1&2z+2\end{vmatrix}\\&=2\begin{vmatrix}x^2&2x+1&1\\y^2&2y+1&1\\ z^2&2z+1&1\end{vmatrix}\\&=2\begin{vmatrix}x^2&2x&1\\y^2&2y&1\\z^2&2z&1\end{vmatrix}\\&=4\begin{vmatrix}x^2&x&1\\y^2&y&1\\z^2&z&1\end{vmatrix}\\&=4(x^2 y-x^2 z-x y^2+x z^2+y^2 z-y z^2)\\&=4(x-y)(x y-x z-y z+z^2)\\&=4(x-y)(x-z)(y-z).\end{align}

José Carlos Santos
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