Find the sum $ \sum\limits_{a,b=0}^ \infty \begin{vmatrix} x^{a+b} & y^{a+b} & z^{a+b}\\ x^b & y^b & z^b \\ 1 & 1 & 1 \end{vmatrix} u^a v^b. $ Find a close determinant expression for the sum $$ \sum_{a,b=0}^ \infty \begin{vmatrix}
x^{a+b} & y^{a+b} & z^{a+b}\\
x^b & y^b & z^b \\ 
1 & 1 & 1
\end{vmatrix} u^a v^b.
$$
Is is not too hard expand the determinant and get that
\begin{gather*}
\sum_{a,b=0}^ \infty \begin{vmatrix}
x^{a+b} & y^{a+b} & z^{a+b}\\
x^b & y^b & z^b \\ 
1 & 1 & 1
\end{vmatrix} u^a v^b=\\=\sum_{a,b=0}^ \infty ({x}^{a+b}{y}^{b}+{y}^{a+b}{z}^{b}-{x}^{a+b}{z}^{b}-{
x}^{b}{y}^{a+b}+{x}^{b}{z}^{a+b}-{z}^{a+b}{y}^{b}) u^a v^b=
\\=
{\frac {1}{
 \left( xu-1 \right)  \left( vxy-1 \right) }}+{\frac {1}{ \left( yu-1
 \right)  \left( vyz-1 \right) }}-{\frac {1}{ \left( xu-1 \right) 
 \left( vxz-1 \right) }}-\\-{\frac {1}{ \left( yu-1 \right)  \left( vxy-1
 \right) }}+{\frac {1}{ \left( zu-1 \right)  \left( vxz-1 \right) }}-{\frac {1}{ \left( zu-1 \right)  \left( vyz-1 \right) }}=\\=
{\frac {uv \left( y-z \right)  \left( x-z \right)  \left( x-y
 \right)  \left( uvxyz-1\right) }{ \left( zu-1 \right)  \left( vyz-1
 \right)  \left( xu-1 \right)  \left( vxy-1 \right)  \left( yu-1
 \right)  \left( vxz-1 \right) }}
\end{gather*}
Question.  Is it possible to go back to determinant and express the result
$$
{\frac {uv \left( y-z \right)  \left( x-z \right)  \left( x-y
 \right)  \left( uvxyz-1\right) }{ \left( zu-1 \right)  \left( vyz-1
 \right)  \left( xu-1 \right)  \left( vxy-1 \right)  \left( yu-1
 \right)  \left( vxz-1 \right) }}
$$
again as determinant ( or sum ( product) of determinants)?
Ant help?

$$
\sum_{a,b=0}^ \infty \begin{vmatrix}
x^{a+b} & y^{a+b} & z^{a+b}\\
x^b & y^b & z^b \\ 
1 & 1 & 1
\end{vmatrix} u^a v^b = \\
\sum_{a,b=0}^ \infty \begin{vmatrix}
u^ax^{a+b} & u^ay^{a+b} & u^az^{a+b}\\
x^b & y^b & z^b \\ 
1 & 1 & 1
\end{vmatrix}  v^b =\\
\sum_{b=0}^ \infty \begin{vmatrix}
\frac{1}{1-xu}x^{b} & \frac{1}{1-yu}y^{b} & \frac{1}{1-zu}z^{b}\\
x^b & y^b & z^b \\ 
1 & 1 & 1
\end{vmatrix}  v^b =\\
\sum_{b=0}^ \infty \begin{vmatrix}
\frac{1}{1-xu} & \frac{1}{1-yu} & \frac{1}{1-zu}\\
1 & 1 & 1 \\ 
x^{-b} & y^{-b} & z^{-b}
\end{vmatrix}  (vxyz)^b =\\
\sum_{b=0}^ \infty \begin{vmatrix}
\frac{1}{1-xu} & \frac{1}{1-yu} & \frac{1}{1-zu}\\
1 & 1 & 1 \\ 
(yzv)^{b} & (xzv)^{b} & (xyv)^{b}
\end{vmatrix}   =\\
\begin{vmatrix}
\frac{1}{1-xu} & \frac{1}{1-yu} & \frac{1}{1-zu}\\
1 & 1 & 1 \\ 
\frac{1}{1-yzv}& \frac{1}{1-xzv} & \frac{1}{1-xyv}
\end{vmatrix} 
$$
