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