Geometric reason why this determinant can be factored to (x-y)(y-z)(z-x)? The determinant $\begin{vmatrix}
    1 & 1 &1 \\ 
    x & y & z \\
    x^2 & y^2 &z^2 \\
  \end{vmatrix}$ can be factored to the form $(x-y)(y-z)(z-x)$

Proof:
Subtracting column 1 from column 2, and putting that in column  2,
\begin{equation*}
\begin{vmatrix}
1 & 1 &1 \\ 
x & y & z \\
x^2 & y^2 &z^2 \\
\end{vmatrix} 
= 
\begin{vmatrix}
1 & 0 &1 \\ 
x & y-x & z \\
x^2 & y^2-x^2 &z^2 \\
\end{vmatrix} 
\end{equation*}
$
= z^2(y-x)-z(y^2-x^2)+x(y^2-x^2)-x^2(y-x)
$
rearranging the terms,
$
=z^2(y-x)-x^2(y-x)+x(y^2-x^2)-z(y^2-x^2)
$
taking out the common terms $(y-x)$ and $(y^2-x^2)$,
$
=(y-x)(z^2-x^2)+(y^2-x^2)(x-z)
$
expanding the terms $(z^2-x^2)$ and $(y^2-x^2)$
$
=(y-x)(z-x)(z+x)+(y-x)(y+x)(x-z)
$
$
=(y-x)(z-x)(z+x)-(y-x)(z-x)(y+x)
$
taking out the common term (y-x)(z-x)
$
=(y-x)(z-x) [z+x-y-x]
$
$
=(y-x)(z-x)(z-y)
$
$
=(x-y)(y-z)(z-x)
$

Is there a geometric reason for this?
The determinant of this matrix is the volume of a parallelopiped with sides as vectors whose tail is at the origin and head at x,y,z coordinates being equal to the columns(or rows) of the matrix.$^{[1]}$
So is the volume of this parallelopiped equals $(x-y)(y-z)(z-x)$ in any obvious geometric way?

References
[1] Nykamp DQ, “The relationship between determinants and area or volume.” From Math Insight. http://mathinsight.org/relationship_determinants_area_volume

Subtracting a multiple of another column (or row) to an
existing column (or row) does not change the determinant.
$$
\begin{vmatrix}
1 & 1 & 1 \\ 
x & y & z \\
x^2 & y^2 & z^2 \\
\end{vmatrix} $$
$$= 
\begin{vmatrix}
1 & 0 &0 \\ 
x & y-x & z-x \\
x^2 & y^2-x^2 &z^2-x^2 \\
\end{vmatrix}$$
$$= 
\begin{vmatrix}
1 & 0 &0 \\ 
0 & y-x & z-x \\
0 & y^2-x^2 &z^2-x^2 \\
\end{vmatrix}$$
$$={(y-x)(z-x) 
\begin{vmatrix}
1 & 0 & 0 \\ 
0 & 1 & 1 \\
0 & y+x &z+x \\
\end{vmatrix} }
$$
$$=(y-x)(z-x) 
\begin{vmatrix}
1 & 0 & 0 \\ 
0 & 1 & 0 \\
0 & y+x & z-y \\
\end{vmatrix} 
$$
$$=(y-x)(z-x)(z-y)$$
