Factorise a matrix using the factor theorem Can someone check this please?
 $$
\begin{vmatrix}
x&y&z\\
x^2&y^2&z^2\\
x^3&y^3&z^3\\
\end{vmatrix}$$
$$C_2=C_2-C_1\implies\quad
\begin{vmatrix}
x&y-x&z\\
x^2&y^2-x^2&z^2\\
x^3&y^3-x^3&z^3\\
\end{vmatrix}$$
$$(y-x)
\begin{vmatrix}
x&1&z\\
x^2&y+x&z^2\\
x^3&y^2+xy+x^2&z^3\\
\end{vmatrix}$$
$$(y-x)(z-x)
\begin{vmatrix}
x&1&1\\
x^2&y+x&z+x\\
x^3&y^2+xy+x^2&z^2+xz+x^2\\
\end{vmatrix}$$
$$R_2=R_2-xR_1\implies\quad
(y-x)(z-x)
\begin{vmatrix}
x&1&1\\
0&y&z\\
x^3&y^2+xy+x^2&z^2+xz+x^2\\
\end{vmatrix}$$
$$R_3=R_3-x^2R_1\implies\quad
(y-x)(z-x)
\begin{vmatrix}
x&1&1\\
0&y&z\\
0&y^2+xy&z^2+xz\\
\end{vmatrix}$$
factor $x$$$\implies\quad
x(y-x)(z-x)
\begin{vmatrix}
1&1&1\\
0&y&z\\
0&y^2+xy&z^2+xz\\
\end{vmatrix}$$
$$\implies\quad x(y-x)(z-x)(yz^2-zy^2)$$
$$\implies\quad xyz(y-x)(z-x)(z-y)$$
Also I'd like practical tips on using the factor theorem for these types of questions. My understanding is that the determinant is $f(x,y,z)$ so if we hold $y$ and $z$ constant we could apply it somehow to $f(x)$ alone. I'm not that great spotting difference of squares etc and want a more fail safe alternative. Thanks in advance.

you are not factoring the matrix. you are using the properties of the determinants to simplify. for example, you could write your first step as 
$$AE = \pmatrix{x & y & z \\ x^2 & y^2 & z^2 \\ x^3 & y^3 & z^3} 
\pmatrix{1 & -1 & 0\\0&1&0\\0&0&1} =
\pmatrix{x & y-x & z \\ x^2 & y^2-x^2 & z^2 \\ x^3 & y^3 -x^3& z^3} = B
 $$
the $E$ matrices are called elementary column matrices and their determinants usually is the product of the diagonals. now use the product rule of the determinants $det(AE) = det(A) det(E)$ to conclude that $det(A) = det(B)$ now you start with $B$ and do further reductions.
