<s>[INST] Calculate the determinant $\left|\begin{smallmatrix} a&b&c&d\\ b&a&d&c\\ c&d&a&b\\d&c&b&a\end{smallmatrix}\right|$ Question: Calculate the following determinant
$$A=\begin{vmatrix} a&b&c&d\\ b&a&d&c\\ c&d&a&b\\d&c&b&a\end{vmatrix}$$
Progress: So I apply $R1'=R1+R2+R3+R4$ and get
$$A=(a+b+c+d)\begin{vmatrix} 1&1&1&1\\ b&a&d&c\\ c&d&a&b\\d&c&b&a\end{vmatrix}$$
Then, I apply $C2'=C2-C1,\, C3'=C3-C1$, etc to get
$$A=(a+b+c+d)\begin{vmatrix}1&0&0&0\\b&a-b&d-b&c-b\\c&d-c&a-c&b-c\\d&c-d&b-d&a-d \end{vmatrix}$$
Thus, $$A=(a+b+c+d)\begin{vmatrix}a-b&d-b&c-b\\d-c&a-c&b-c\\c-d&b-d&a-d \end{vmatrix}$$
But now I'm stuck here. I don't really want to expand the whole thing, because it is really messy. Is there any approach that don't require much calculation?
 [/INST] \begin{align}A&=(a+b+c+d)\begin{vmatrix}a-b&d-b&c-b\\d-c&a-c&b-c\\c-d&b-d&a-d \end{vmatrix} \\&=(a+b+c+d)\begin{vmatrix}a-b&d-b&c-b\\d-c&a-c&b-c\\0 &a+b-c-d&a+b-c-d \end{vmatrix} \\
&= (a+b+c+d)(a+b-c-d)\begin{vmatrix}a-b&d-b&c-b\\d-c&a-c&b-c\\0 &1&1\end{vmatrix} \\
&= (a+b+c+d)(a+b-c-d)\begin{vmatrix}a-b&d-b&c-b\\a+d-b-c&a+d-b-c& 0\\0 &1&1\end{vmatrix} \\
&=(a+b+c+d)(a+b-c-d)(a+d-b-c)\begin{vmatrix}a-b&d-b&c-b\\1&1& 0\\0 &1&1\end{vmatrix} \\
&=(a+b+c+d)(a+b-c-d)(a+d-b-c)\begin{vmatrix}a-b&d-c&0\\1&1& 0\\0 &1&1\end{vmatrix} \\
&=(a+b+c+d)(a+b-c-d)(a+d-b-c)(a-b+c-d)
\end{align}
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