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

Let $R_1, R_2, R_3$ be the first, second, and third rows.
Apply the operations $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$.

$$\begin{vmatrix} 1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c \end{vmatrix} = \begin{vmatrix} 1+a & 1 & 1 \\ 1-(1+a) & (1+b)-1 & 1-1 \\ 1-(1+a) & 1-1 & (1+c)-1 \end{vmatrix}$$ $$= \begin{vmatrix} 1+a & 1 & 1 \\ -a & b & 0 \\ -a & 0 & c \end{vmatrix}$$

$$= (1+a) \begin{vmatrix} b & 0 \\ 0 & c \end{vmatrix} - (-a) \begin{vmatrix} 1 & 1 \\ 0 & c \end{vmatrix} + (-a) \begin{vmatrix} 1 & 1 \\ b & 0 \end{vmatrix}$$

Calculate the $2 \times 2$ determinants:

$$\begin{vmatrix} b & 0 \\ 0 & c \end{vmatrix} = (b)(c) - (0)(0) = bc$$ $$\begin{vmatrix} 1 & 1 \\ 0 & c \end{vmatrix} = (1)(c) - (1)(0) = c$$ $$\begin{vmatrix} 1 & 1 \\ b & 0 \end{vmatrix} = (1)(0) - (1)(b) = -b$$

$$= (1+a)(bc) - (-a)(c) + (-a)(-b)$$ $$= bc + abc + ac + ab$$

Rearrange the terms to match the target expression:

$$= abc + ab + bc + ca$$

Thus, we have shown that:

$$\boxed{\begin{vmatrix} 1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c \end{vmatrix} = abc+ab+bc+ca}$$