I'm learning how to factorize determinants of a square matrix in school, but we haven't learnt a general method to do that, besides 'creating zeros'. So I thought maybe I'll ask here if someone does know a method that generalizes the factorization of a square matrix.
An example
I had this determinant to factorize on a test:
$\left| \begin{array}{ccc} x & y & 1 \\ x^2 & y^2 & 1 \\ x^3 & y^3 & 1 \end{array} \right| $
So I started and got the following, quite easy:
$ \begin{align} \left| \begin{array}{ccc} x & y & 1 \\ x^2 & y^2 & 1 \\ x^3 & y^3 & 1 \end{array} \right| &= xy \cdot \left| \begin{array}{ccc} 1 & 1 & 1 \\ x & y & 1 \\ x^2 & y^2 & 1 \\ \end{array} \right| & &(1) \end{align} $
At this point I was stuck. I had now idea how to go on from this point. But when I got my test back, it was corrected by the teacher:
$ \begin{align*} xy \cdot \left| \begin{array}{ccc} 1 & 1 & 1 \\ x & y & 1 \\ x^2 & y^2 & 1 \\ \end{array} \right| &= xy \cdot \left| \begin{array}{ccc} 0 & 0 & 1 \\ x - 1 & y -1 & 1 \\ x^2 -1 & y^2 - 1 & 1 \\ \end{array} \right| \\ &= xy \cdot (x-1)(y-1) \left| \begin{array}{ccc} 0 & 0 & 1 \\ 1 & 1 & 1 \\ x + 1 & y + 1 & 1 \end{array} \right| & \text{because } x^2 - 1^2 = (x-1)(x+1)\\ &=xy \cdot (x-1)(y-1) \cdot 1 \cdot(1 \cdot (y+1) - 1 \cdot (x + 1)) & \text{Laplace with row 1}\\ &= xy \cdot (x-1)(x+1)(y-x) \end{align*} $
Alright, so I had to see myself at (1) that I had to subtract row 3 from row 1 and 2. Then I had to see I had to use the formula $a^2 - b^2 = (a-b)(a+b)$. I'm guessing a lot of you didn't see you had to do that. So my question to you: is there an easier way?
PS: I'll accept 'no' for an answer!