Prove that:

Let LHS $=\Delta=\mid \begin{array}{lll}a & b-c & c-b\end{array}$

$\begin{array}{clc}a-c & b & c-a \\ a-b & b-a & c \mid\end{array}$

$\Delta=\mid a \quad 0 \quad c-b+a$

$\begin{array}{ccc}a-c & b+c-a & 0 \\ a-b & b+c-a & c+a-b \mid\end{array}$        [Applying $\mathrm{C}_{2} \rightarrow \mathrm{C}_{2}+\mathrm{C}_{3}$ and $\mathrm{C}_{3} \rightarrow \mathrm{C}_{1}+\mathrm{C}_{3}$ ]

$=(b+c-a)(c+a-b) \mid a \quad 0 \quad 1$

$\begin{array}{llll}a-c & 1 & 0 & \\ a-b & 1 & 1 & \text { [Taking out common factor from } \mathrm{C}_{2} \text { and } \mathrm{C}_{3} \text { ] }\end{array}$

$1 \quad 1 \mid)+(1 \times \quad \mid a-c \quad 1$

$\left.\begin{array}{ll}a-b & 1 \mid)\end{array}\right\} \quad\left[\right.$ Expanding along $\left.R_{1}\right]$

$=(a+b-c)(b+c-a)(c+a-b)$

$=$ RHS