Question

Prove that :

\[\begin{vmatrix}1 & b + c & b^2 + c^2 \\ 1 & c + a & c^2 + a^2 \\ 1 & a + b & a^2 + b^2\end{vmatrix} = \left( a - b \right) \left( b - c \right) \left( c - a \right)\]

 

Answer 1

\[\text{ Let LHS }= ∆ = \begin{vmatrix} 1 & b + c & b^2 + c^2 \\1 & c + a & c^2 + a^2 \\1 & a + b & a^2 + b^2 \end{vmatrix}\]

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