General method for factorizing matrix determinants 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!

There is generally no easy way. Concise (closed) formula for the determinants of a general matrix is rare. For your example, it is too special, the determiant is equal to $\left| 
\begin{array}{cccc}1 & 1 & 1 & 1\\
x & y & 1 & 0\\
x^2 & y^2 & 1 &0\\
x^3 & y^3 & 1&0
\end{array} 
\right|$, where the matrix is a Vandermonde matrix whose determiant has a closed expression.
