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प्रश्न
Prove that :
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उत्तर
\[\text{ Let LHS }= \Delta = \begin{vmatrix} a^2 & a^2 - \left( b - c \right)^2 & bc\\ b^2 & b^2 - \left( c - a \right)^2 & ca\\ c^2 & c^2 - \left( a - b \right)^2 & ab \end{vmatrix}\]
\[ \Rightarrow ∆ = \begin{vmatrix} a^2 & - \left( b - c \right)^2 & bc\\ b^2 & - \left( c - a \right)^2 & ca\\ c^2 & - \left( a - b \right)^2 & ab \end{vmatrix} \left[\text{ Applying }C_2 \to C_2 - C_1 \right]\]
\[ = \left( - 1 \right)\begin{vmatrix} a^2 & \left( b - c \right)^2 & bc\\ b^2 & \left( c - a \right)^2 & ca\\ c^2 & \left( a - b \right)^2 & ab \end{vmatrix}\]
\[ = - \begin{vmatrix} a^2 & b^2 + c^2 & bc\\ b^2 & c^2 + a^2 & ca\\ c^2 & a^2 + b^2 & ab \end{vmatrix} \left[\text{ Applying }C_2 \to C_2 - 2 C_3 \right]\]
\[ = - \begin{vmatrix} a^2 + b^2 + c^2 & b^2 + c^2 & bc\\ b^2 + c^2 + a^2 & c^2 + a^2 & ca\\ c^2 + a^2 + b^2 & a^2 + b^2 & ab \end{vmatrix} \left[\text{ Applying }C_1 \to C_1 + C_2 \right]\]
\[ = - \left( a^2 + b^2 + c^2 \right)\begin{vmatrix} 1 & b^2 + c^2 & bc\\1 & c^2 + a^2 & ca\\1 & a^2 + b^2 & ab \end{vmatrix}\]
\[ = - \left( a^2 + b^2 + c^2 \right)\begin{vmatrix} 1 & b^2 + c^2 & bc\\0 & \left( c^2 + a^2 \right) - \left( b^2 + c^2 \right) & ca - bc\\0 & \left( a^2 + b^2 \right) - \left( b^2 + c^2 \right) & ab - bc \end{vmatrix} \left[\text{ Applying }R_2 \to R_2 - R_1 \text{ and }R_3 \to R_3 - R_1 \right]\]
\[ = \left( \left( a^2 + b^2 + c^2 \right) \right)\begin{vmatrix} 1 & b^2 + c^2 & bc\\0 & a^2 - b^2 & c\left( a - b \right)\\0 & a^2 - c^2 & b \left( a - c \right) \end{vmatrix}\]
\[ = - \left( a^2 + b^2 + c^2 \right)\left( a - b^{} \right)\left( a - c \right)\begin{vmatrix} 1 & b^2 + c^2 & bc\\0 & a + b^{} & c\\0 & a^{} + c^{} & b \end{vmatrix} \left[\text{ Taking }\left( a - b \right)\text{ common from }R_2\text{ and }\left( a - c \right)\text{ common from }R_3 \right]\]
\[ = \left( a^2 + b^2 + c^2 \right)\left( a - b^{} \right)\left( c - a \right) \times \left\{ 1 \times \begin{vmatrix} a + b^{} & c\\ a^{} + c^{} & b \end{vmatrix} \right\} \left[ \because \left( c - a \right) = - \left( a - c \right) \right] \left[\text{ Expanding along }C_1 \right]\]
\[ = \left( a^2 + b^2 + c^2 \right)\left( a - b^{} \right) \left( c - a \right) \left( ab + b^2 - ac - c^2 \right) \]
\[= \left( a^2 + b^2 + c^2 \right)\left( a - b \right)\left( c - a \right)\left\{ a\left( b - c \right) + \left( b + c \right)\left( b - c \right) \right\}\]
\[ = \left( a - b \right)\left( c - a \right)\left( b - c \right)\left( a + b + c \right)\left( a^2 + b^2 + c^2 \right)\]
= RHS
Hence proved
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