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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?

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    $\begingroup$ Little typo (I think...) for the northeastern entry. $\endgroup$ – Jean Marie Jun 17 '17 at 20:45
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It's obvious that we have a factor $(a-b)(b-c)(c-a)(x-y)(y-z)(z-x)$

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  • $\begingroup$ yes but the determinant which is multiplied with $f(a,b,c,x,y,z)$ is of $12$th degree right. So with the above factors we get only $6$th degree. What about the other $6$th degree expression? $\endgroup$ – Ekaveera Kumar Sharma Jun 17 '17 at 16:43
  • $\begingroup$ @Ekaveera Kumar Sharma The rest should be something like this: $\sum\limits_{0\leq i,j,k,m,n,l\leq2}a^ib^jc^kx^my^nz^l$, where $i+j+k+m+n+l=6$. $\endgroup$ – Michael Rozenberg Jun 17 '17 at 16:50

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