Cofactor expansion method for finding the determinant of a matrix Use the determinant properties to simplify the given matrix and show that $\det(A) = (x -
y)(x - z)(x - w)(y - z)(y - w)(z - w)$ for
$$A =
\begin{pmatrix}
1 & x & x^2 & x^3 \\
1 & y & y^2 & y^3 \\
1 & z & z^2 & z^3\\
1 & w & w^2 & w^3 \\
\end{pmatrix}
$$
For this problem, I noticed that the area of interest is the second column of matrix $A$. Will using the cofactor expansion method of finding the determinant yield the desired result if I choose the second column? The second column has $x$,$y$,$z$,$w$ and what I am looking for is $\det(A) = (x-y)(x-z)(x-w)(y-z)(y-w)(z-w)$. Thanks for any help!
 A: Here is a non-clever approach that is still better than brute force. Subtract the bottom row from each of the preceding rows. This does not change the determinant. The result is 
$$\begin{pmatrix}
0 & x-w & x^2-w^2 & x^3-w^3 \\
0 & y-w & y^2-w^2 & y^3-w^3 \\
0 & z-w & z^2-w^2 & z^3-w^3 \\
1 & w & w^2 & w^3 
\end{pmatrix}$$
Expanding along the first column,   find that the determinant is 
$$-\det \begin{pmatrix}
  x-w & x^2-w^2 & x^3-w^3 \\
  y-w & y^2-w^2 & y^3-w^3 \\
  z-w & z^2-w^2 & z^3-w^3 
\end{pmatrix}$$
Factor out $(x-w)(y-w)(z-w)$:
$$-(x-w)(y-w)(z-w) \det \begin{pmatrix}
  1 & x+w & x^2+xw+w^2 \\
  1 & y+w & y^2+yw+w^2 \\
  1 & z+w & z^2+zw+w^2 
\end{pmatrix}$$
Again subtract the bottom row from the first two.
$$-(x-w)(y-w)(z-w) \det \begin{pmatrix}
  0 & x-z & x^2-z^2+(x-z)w  \\
  0 & y-z & y^2-z^2+(y-z)w  \\
  1 & z+w & z^2+zw+w^2 
\end{pmatrix}$$
This becomes a $2\times 2$ determinant, from which you can factor
$(x-z)(y-z)$:
$$-(x-w)(y-w)(z-w)(x-z)(y-z) \det \begin{pmatrix}
  1 & x+z+1  \\
  1 & y+z+1   
\end{pmatrix}$$
and conclude.
