Question

\[\begin{vmatrix}a + b + c & - c & - b \\ - c & a + b + c & - a \\ - b & - a & a + b + c\end{vmatrix} = 2\left( a + b \right) \left( b + c \right) \left( c + a \right)\]

Answer 1

\[\text{ Let LHS }= ∆ = \begin{vmatrix} a + b + c & - c & - b\\ - c & a + b + c & - a\\ - b & - a & a + b + c \end{vmatrix}\] 
\[ = \begin{vmatrix} a & - c & - b\\b & a + b + c & - a\\c & - a & a + b + c \end{vmatrix} \left[\text{ Applying }C_1 \to C_1 + C_2 + C_3 \right]\] 
\[ = \begin{vmatrix} a + b & a + b & - \left( a + b \right) \\b + c & b + c & b + c \\c & - a & a + b + c \end{vmatrix} \left[\text{ Applying }R_1 \to R_1 + R_2\text{ and }R_2 \to R_2 + R_3 \right]\] 
\[ = \left( a + b \right)\left( b + c \right) \begin{vmatrix} 1 & 1 & - 1 \\ 1 & 1 & 1\\ c & - a & a + b + c \end{vmatrix} \left[\text{ Taking out common factor from }R {}_1\text{ and }R_2 \right]\] 
\[ = \left( a + b \right)\left( b + c \right)\begin{vmatrix} 0 & 0 & - 2\\ 1 & 1 & 1 \\ c & - a & a + b + c \end{vmatrix} \left[\text{ Applying }R_1 \to R_1 - R_2 \right]\] 
\[ = \left( a + b \right)\left( b + c \right)\left\{ \left( - 2 \right)\left( - a - c \right) \right\} \left[\text{ Expanding along }R_1 \right]\] 
\[ = 2 \left( a + b \right)\left( b + c \right) \left( c + a \right) \] 
\[ = RHS\]
Hence proved.