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 $(a-b)(b-c)(c-a)(x-y)(y-z)(z-x)$