Evaluate the following Determinant of $12$th degree polynomial Evaluate $$\Delta=\begin{vmatrix}
\frac{1}{(a+x)^2} & \frac{1}{(b+x)^2} & \frac{1}{(c+x^2)}\\ 
\frac{1}{(a+y)^2} & \frac{1}{(b+y)^2} & \frac{1}{(c+y)^2}\\ 
\frac{1}{(a+z)^2} & \frac{1}{(b+z)^2} & \frac{1}{(c+z)^2}\\ 
\end{vmatrix}$$
My Try: I have taken all the denominators out and we obtain
$$\Delta=f(a,b,c,x,y,z)\times \begin{vmatrix}
(b+x)^2(c+x)^2 & (a+x)^2(c+x)^2 & (a+x)^2(b+x)^2\\ 
 (b+y)^2(c+y)^2 & (a+y)^2(c+y)^2 & (a+y)^2(b+y)^2\\ 
 (b+z)^2(c+z)^2 & (a+z)^2(c+z)^2 & (a+z)^2(b+z)^2\\ 
\end{vmatrix}$$  where
$$f(a,b,c,x,y,z)=\frac{1}{\left((a+x)(b+x)(c+x)(a+y)(b+y)(c+y)(a+z)(b+z)(c+z)\right)^2} $$
By factor theorem we observe that $a-b$,$b-c$,$c-a$,$x-y$,$y-z$ and $z-x$ are factors of the new Determinant above.
But how to find remaining factors?

It's obvious that we have a factor $(a-b)(b-c)(c-a)(x-y)(y-z)(z-x)$
