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Question
Prove that :
Solution
\[\text{ Let LHS }= \Delta = \begin{vmatrix} a + b + 2c & a & b\\c & b + c + 2a & b\\c & a & c + a + 2b \end{vmatrix}\]
\[ \Rightarrow ∆ = \begin{vmatrix} 2a + 2b + 2c & a & b \\2a + 2b + 2c & b + c + 2a & b\\2a + 2b + 2c & a & c + a + 2b \end{vmatrix}\left[\text{ Applying }C_1 \to C_1 + C_2 + C_3 \right] \]
\[ = 2 \left( a + b + c \right) \begin{vmatrix} 1 & a & b \\1 & b + c + 2a & b\\1 & a & c + a + 2b \end{vmatrix} \left[\text{ Taking out 2(a + b + c) common from }C_1 \right]\]
\[ ∆ = 2 \left( a + b + c \right) \begin{vmatrix} 1 & a & b \\0 & b + c + a & 0\\0 & - b - c - a & c + a + b \end{vmatrix} \left[\text{ Applying }R_2 \to R_2 - R_1\text{ and }R_2 \to R_2 - R_3 \right]\]
\[ = 2\left( a + b + c \right)\left( a + b + c \right)\left( a + b + c \right)\begin{vmatrix} 1 & a & b \\0 & 1 & 0\\0 & - 1 & 1 \end{vmatrix} \left[\text{ Taking out (a + b + c) common from }R_2\text{ and }R_3 \right]\]
\[ = 2 \left( a + b + c \right)^3 \left\{ 1\left( 1 - 0 \right) \right\} \left[\text{ Expanding along }C_1 \right]\]
\[ = 2 \left( a + b + c \right)^3 \]
\[ = RHS\]
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