Easier way of calculating the determinant for this matrix

You can easily transform your matrix to a circulant matrix $$ M=\left(\begin{array}{cccc} a&b&c&d\\ d&a&b&c\\ c&d&a&b\\ b&c&d&a \end{array}\right) $$ by carrying out obvious row swaps. The eigenvalue theory of circulant matrices is completely known. In the $4\times 4$ case the eigenvalues of $M$ are $$ \lambda_1=a+b+c+d,\ \lambda_2=a+bi-c-di,\ \lambda_3=a-b+c-d\ \text{and}\ \lambda_4=a-bi-c+di. $$ The determinant of a matrix is the product of its eigenvalues so $$ \det M=\lambda_1\lambda_2\lambda_3\lambda_4. $$

A pedestrian's solution (experimentX's suggestion below the question).

Add the first three columns to the fourth: \begin{align*} \begin{vmatrix} a&b&c&d\\b&c&d&a\\c&d&a&b\\d&a&b&c \end{vmatrix} & =(a+b+c+d)\begin{vmatrix} a&b&c&1\\b&c&d&1\\c&d&a&1\\d&a&b&1 \end{vmatrix} \end{align*} Subtract the second row from the first row, the third row from the second row and the fourth row from the third row; develop after the fourth column:

\begin{align*} &=(a+b+c+d) \begin{vmatrix} a-b&b-c&c-d \\ b-c&c-d&d-a \\ c-d&d-a&a-b \end{vmatrix} \\ \end{align*}

Add the first column to the third column:

\begin{align*} &=(a+b+c+d)(a-b+c-d) \begin{vmatrix} a-b&b-c&1 \\ b-c&c-d&-1 \\ c-d&d-a&1 \end{vmatrix} \end{align*} Add the second row to the first row and the third row to the second row: \begin{align*} &=(a+b+c+d)(a-b+c-d) \begin{vmatrix} a-c&b-d&0 \\ b-d&c-a&0 \\ c-d&d-a&1 \end{vmatrix} \\ &=-(a+b+c+d)(a-b+c-d)[(a-c)^2+(b-d)^2]. \end{align*}