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Prove that : ∣ ∣ ∣ ∣ ∣ ( B + C ) 2 a 2 B C ( C + a ) 2 B 2 C a ( a + B ) 2 C 2 a B ∣ ∣ ∣ ∣ ∣ = ( a − B ) ( B − C ) ( C − a ) ( a + B + C ) ( a 2 + B 2 + C 2 )

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Question

Prove that :

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

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Solution

\[\text{ Let LHS }= \Delta = \begin{vmatrix} \left( b + c \right)^2 & a^2 & bc\\ \left( c + a \right)^2 & b^2 & ca\\ \left( a + b \right)^2 & c^2 & ab \end{vmatrix}\] 
\[ = \begin{vmatrix} \left( b + c \right)^2 - \left( c + a \right)^2 & a^2 - b^2 & bc - ca\\ \left( c + a \right)^2 - \left( a + b \right)^2 & b^2 - c^2 & ca - ab\\ \left( a + b \right)^2 & c^2 & ab \end{vmatrix} \left[\text{ Applying }R_1 \to R_1 - R_2\text{ and }R_2 \to R_2 - R_3 \right]\] 
\[ = \begin{vmatrix} \left( b - a \right)\left( b + 2c + a \right) & \left( a + b \right) \left( a - b \right)b & c\left( b - a \right)\\ \left( c - b \right)\left( b + 2a + c \right) & \left( b - c \right) \left( b + c \right) & a\left( c - b \right)\\ \left( a + b \right)^2 & c^2 & ab \end{vmatrix}\] 
\[ = \left( a - b \right)\left( b - c \right)\begin{vmatrix} - \left( b + 2c + a \right) & a + b & - c \\ - \left( b + 2a + c \right) & b + c & - a\\ \left( a + b \right)^2 & c^2 & ab \end{vmatrix} \left[\text{ Applying }x^2 - y^2 = \left( x + y \right)\left( x - y \right)\text{ and taking out }\left( a - b \right)\text{ common from }R_1\text{ and }\left( b - c \right)\text{ from }R_2 \right]\] 
\[ = \left( a - b \right)\left( b - c \right) \begin{vmatrix} - 2\left( b + c + a \right) & a + b & - c \\ - 2\left( b + a + c \right) & b + c & - a\\ \left( a + b \right)^2 - c^2 & c^2 & ab \end{vmatrix} \left[\text{ Applying }C_1 \to C_1 - C_2 \right]\]
\[ = \left( a - b \right)\left( b - c \right) \begin{vmatrix} - 2\left( b + c + a \right) & a + b & - c \\ - 2\left( b + a + c \right) & b + c & - a\\ \left( a + b + c \right) \left( a + b - c \right) & c^2 & ab \end{vmatrix} \left[\text{ Applying }x^2 - y^2 = \left( x + y \right)\left( x - y \right)\text{ in }C_1 \right]\] 
\[ = \left( a - b \right)\left( b - c \right)\left( a + b + c \right) \begin{vmatrix} - 2 & a + b & - c \\ - 2 & b + c & - a\\ \left( a + b - c \right) & c^2 & ab \end{vmatrix} \left[\text{ Taking out }\left( a + b + c \right)\text{ common from }C_1 \right]\] 
\[ = \left( a - b \right)\left( b - c \right)\left( a + b + c \right)\begin{vmatrix} - 2 & a + b & - c \\ 0 & c - a & c - a\\ \left( a + b - c \right) & c^2 & ab \end{vmatrix} \left[\text{ Applying }R_2 \to R_2 - R_1 \right]\] 
\[ = \left( a - b \right)\left( b - c \right)\left( a + b + c \right)\left( c - a \right)\begin{vmatrix} - 2 & a + b & - c \\ 0 & 1 & 1\\\left( a + b - c \right) & c^2 & ab \end{vmatrix} \left[\text{ Taking out }\left( c - a \right)\text{ common from }R_2 \right]\] 
\[ = \left( a - b \right)\left( b - c \right)\left( a + b + c \right)\left( c - a \right)\begin{vmatrix} - 2 & a + b + c & - c \\ 0 & 0 & 1\\\left( a + b - c \right) & c^2 - ab & ab \end{vmatrix} \left[\text{ Applying }C_2 \to C_2 - C_3 \right]\] 
\[ = \left( a - b \right)\left( b - c \right)\left( a + b + c \right)\left( c - a \right) \left\{ \left( - 1 \right)\begin{vmatrix} - 2 & a + b + c \\\left( a + b - c \right) & c^2 - ab \end{vmatrix} \right\} \left[\text{ Expanding along }R_2 \right]\] 
\[ = - \left( a - b \right)\left( b - c \right)\left( a + b + c \right)\left( c - a \right)\left\{ - 2 c^2 + 2ab - a^2 - b^2 - 2ab + c^2 \right\}\] 
\[ = - \left( a - b \right)\left( b - c \right)\left( a + b + c \right)\left( c - a \right)\left( - a^2 - b^2 - c^2 \right)\] 
\[ = \left( a - b \right)\left( b - c \right)\left( a + b + c \right)\left( c - a \right)\left( a^2 + b^2 + c^2 \right)\]
\[ = RHS\] 

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Chapter 5: Determinants - Exercise 6.2 [Page 59]

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R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 5 Determinants
Exercise 6.2 | Q 20 | Page 59

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