Show that $\left|\begin{array}{ccc}1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c\end{array}\right|=abc\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=abc+bc+ca+ab$.

(A) Taking out factors $a, b, c$ common from $R_{1}, R_{2}$ and $R_{3}$ respectively,we get:
$\text{L.H.S.} = abc \left|\begin{array}{ccc} \frac{1}{a}+1 & \frac{1}{a} & \frac{1}{a} \\ \frac{1}{b} & \frac{1}{b}+1 & \frac{1}{b} \\ \frac{1}{c} & \frac{1}{c} & \frac{1}{c}+1 \end{array}\right|$
Applying $R_{1} \rightarrow R_{1}+R_{2}+R_{3}$,we have:
$\Delta = abc \left|\begin{array}{ccc} 1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c} & 1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c} & 1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \\ \frac{1}{b} & \frac{1}{b}+1 & \frac{1}{b} \\ \frac{1}{c} & \frac{1}{c} & \frac{1}{c}+1 \end{array}\right|$
$= abc \left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \left|\begin{array}{ccc} 1 & 1 & 1 \\ \frac{1}{b} & \frac{1}{b}+1 & \frac{1}{b} \\ \frac{1}{c} & \frac{1}{c} & \frac{1}{c}+1 \end{array}\right|$
Now applying $C_{2} \rightarrow C_{2}-C_{1}$ and $C_{3} \rightarrow C_{3}-C_{1}$,we get:
$\Delta = abc \left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \left|\begin{array}{ccc} 1 & 0 & 0 \\ \frac{1}{b} & 1 & 0 \\ \frac{1}{c} & 0 & 1 \end{array}\right|$
$= abc \left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) [1(1-0)]$
$= abc \left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) = abc + bc + ca + ab = \text{R.H.S.}$