Show that |1+a 1 1| |1 1+b 1| |1 1 1+c| = abc+ab+bc+ca.

Here is the complete solution for Problem 8. Problem 8: Show that 1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c = abc+ab+bc+ca. Step 1: Apply row operations to simplify the determinant. Let R_1, R_2, R_3 be the first, second, and third rows. Apply the operations R_2 R_2 - R_1 and R_3 R_3 - R_1. 1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c = 1+a & 1 & 1 \\ 1-(1+a) & (1+b)-1 & 1-1 \\ 1-(1+a) & 1-1 & (1+c)-1 = 1+a & 1 & 1 \\ -a & b & 0 \\ -a & 0 & c Step 2: Expand the determinant along the first column. = (1+a) b & 0 \\ 0 & c - (-a) 1 & 1 \\ 0 & c + (-a) 1 & 1 \\ b & 0 Calculate the 2 × 2 determinants: b & 0 \\ 0 & c = (b)(c) - (0)(0) = bc 1 & 1 \\ 0 & c = (1)(c) - (1)(0) = c 1 & 1 \\ b & 0 = (1)(0) - (1)(b) = -b Step 3: Substitute the 2 × 2 determinants back into the expansion and simplify. = (1+a)(bc) - (-a)(c) + (-a)(-b) = bc + abc + ac + ab Rearrange the terms to match the target expression: = abc + ab + bc + ca Thus, we have shown that: 1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c = abc+ab+bc+ca