Evaluate the following Determinant Find $$ \Delta=\begin{vmatrix}
\frac{1}{a+x} &\frac{1}{b+x}  &\frac{1}{c+x} \\ 
 \frac{1}{a+y} &\frac{1}{b+y}  &\frac{1}{c+y} \\ 
\frac{1}{a+z} &\frac{1}{b+z}  &\frac{1}{c+z}
\end{vmatrix}$$
I applied $C_1 \to C_1-C_2$ and $C_2 \to C_2-C_3$ we get
$$ \Delta=\begin{vmatrix}
\frac{b-a}{(a+x)(b+x)} &\frac{c-b}{(b+x)(c+x)}  &\frac{1}{c+x} \\ 
 \frac{b-a}{(a+y)(b+y)} &\frac{c-b}{(b+y)(c+y)}&\frac{1}{c+y} \\ 
\frac{b-a}{(a+z)(b+z)} &\frac{c-b}{(b+z)(c+z)}  &\frac{1}{c+z}
\end{vmatrix}$$
Now taking $b-a$,$\:$$c-b$ common and taking $$\frac{1}{(a+x)(b+x)(c+x)(a+y)(b+y)(c+y)(a+z)(b+z)(c+z)}$$ outside Determinant we get
$$\Delta=\frac{(b-a)(c-b)}{(a+x)(b+x)(c+x)(a+y)(b+y)(c+y)(a+z)(b+z)(c+z)} \times \begin{vmatrix}
c+x &a+x &(a+x)(b+x) \\ 
c+y &a+y &(a+y)(b+y)\\ 
c+z &a+z &(a+z)(b+z)
\end{vmatrix}$$
Now apply $C_1 \to C_1-C_2$ we get
$$\Delta=\frac{(b-a)(c-b)(c-a)}{(a+x)(b+x)(c+x)(a+y)(b+y)(c+y)(a+z)(b+z)(c+z)} \times \begin{vmatrix}
1 &a+x &(a+x)(b+x) \\ 
1 &a+y &(a+y)(b+y)\\ 
1 &a+z &(a+z)(b+z)
\end{vmatrix}$$
any clue  here?

\begin{align}
\Delta&=\begin{vmatrix} \frac{1}{a+x} & \frac{1}{b+x} & \frac{1}{c+x} \\  \frac{1}{a+y} & \frac{1}{b+y} & \frac{1}{c+y} \\ \frac{1}{a+z} & \frac{1}{b+z} & \frac{1}{c+z}\end{vmatrix}\\
\left[\prod_{q\in\{x,y,z\}}(a+q)\right]\Delta&=\begin{vmatrix} 1 & \frac{a+x}{b+x} & \frac{a+x}{c+x} \\  1 & \frac{a+y}{b+y} & \frac{a+y}{c+y} \\ 1 & \frac{a+z}{b+z} & \frac{a+z}{c+z}\end{vmatrix}\\
\left[\prod_{p\in\{a,b\}}\prod_{q\in\{x,y,z\}}(p+q)\right]\Delta&=\begin{vmatrix} b+x & a+x & \frac{(a+x)(b+x)}{c+x} \\  b+y & a+y & \frac{(a+y)(b+y)}{c+y} \\ b+z & a+z & \frac{(a+z)(b+z)}{c+z}\end{vmatrix}\\
\left[\prod_{p\in\{a,b,c\}}\prod_{q\in\{x,y,z\}}(p+q)\right]\Delta&=\begin{vmatrix} (b+x)(c+x) & (c+x)(a+x) & (a+x)(b+x) \\  (b+y)(c+y) & (c+y)(a+y) & (a+y)(b+y) \\ (b+z)(c+z) & (c+z)(a+z) & (a+z)(b+z)\end{vmatrix}
\end{align}
Let $\displaystyle \Delta_0=\begin{vmatrix} (b+x)(c+x) & (c+x)(a+x) & (a+x)(b+x) \\  (b+y)(c+y) & (c+y)(a+y) & (a+y)(b+y) \\ (b+z)(c+z) & (c+z)(a+z) & (a+z)(b+z)\end{vmatrix}$.
Note that $\Delta_0=0$ when $a=b$, $b=c$, $c=a$, $x=y$, $y=z$ and $z=x$.
Since $\Delta_0$ is a polynomial of degree $6$ in $a,b,c,x,y,z$, 
$$\Delta_0=k(a-b)(b-c)(c-a)(x-y)(y-z)(z-x)$$
for some constant $k$.
Put $a=x=1$, $b=y=0$ and $c=z=-1$.
\begin{align}
\begin{vmatrix} 0 & 0 & 2 \\  0 & -1 & 0 \\ 2 & 0 & 0\end{vmatrix}&=k(1)(1)(-2)(1)(1)(-2)\\
k&=1
\end{align}
Therefore,
$$\Delta=\frac{(a-b)(b-c)(c-a)(x-y)(y-z)(z-x)}{(a+x)(a+y)(a+z)(b+x)(b+y)(b+z)(c+x)(c+y)(c+z)}$$
