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