A cyclic 3×3 determinant

Column operation. $C_1\leftarrow C_1+C_2+C_3$:
$$\Delta = \begin{vmatrix} 1+a+b & a & b \\ 1+a+b & 1 & a \\ 1+a+b & b & 1 \end{vmatrix} = (1+a+b)\begin{vmatrix} 1 & a & b \\ 1 & 1 & a \\ 1 & b & 1 \end{vmatrix}.$$

Row operations. $L_2\leftarrow L_2-L_1$, $L_3\leftarrow L_3-L_1$:
$$\Delta = (1+a+b)\begin{vmatrix} 1 & a & b \\ 0 & 1-a & a-b \\ 0 & b-a & 1-b \end{vmatrix}.$$
Expand along the first column:
$$\Delta = (1+a+b)\begin{vmatrix} 1-a & a-b \\ b-a & 1-b \end{vmatrix}.$$

$2\times2$ determinant.
$$\begin{vmatrix} 1-a & a-b \\ b-a & 1-b \end{vmatrix} = (1-a)(1-b) - (a-b)(b-a) = (1-a)(1-b)+(a-b)^2.$$
Expand: $1-a-b+ab+a^2-2ab+b^2 = a^2-ab+b^2-a-b+1$.

Conclusion.
$$\Delta = (1+a+b)(a^2-ab+b^2-a-b+1).$$

Remark. When $a=b=1$, the determinant is $3\cdot(1-1+1-1-1+1)=0$, consistent with a matrix with all entries $1$.