Question

Prove that :

\[\begin{vmatrix}1 & a^2 + bc & a^3 \\ 1 & b^2 + ca & b^3 \\ 1 & c^2 + ab & c^3\end{vmatrix} = - \left( a - b \right) \left( b - c \right) \left( c - a \right) \left( a^2 + b^2 + c^2 \right)\]

 

Answer 1

\[\text{ Let LHS }= \Delta = \begin{vmatrix} 1 & a^2 + bc & a^3 \\1 & b^2 + ca & b^3 \\1 & c^2 + ab & c^3 \end{vmatrix}\] 
\[ \Rightarrow \Delta = \begin{vmatrix} 0 & \left( a^2 + bc \right) - \left( b^2 + ca \right) & a^3 - b^3 \\0 & \left( b^2 + ca \right) - \left( c^2 + ab \right) & b^3 - c^3 \\1 & c^2 + ab & c \end{vmatrix} \left[\text{ Applying }R_1 \to R_1 - R_2\text{ and }R_2 \to R_2 - R_3 \right]\]
\[= \begin{vmatrix} 0 & a^2 - b^2 - ca + bc & a^3 - b^3 \\0 & b^2 - c^2 - ab + ca & b^3 - c^3 \\1 & c^2 + ab & c^3 \end{vmatrix}\] 
\[ = \begin{vmatrix} 0 & \left( a - b \right)  \left( a + b - c \right) &\left( a - b \right)\left( a^2 + ab + b^2 \right)\\0 & \left( b - c \right)\left( b + c - a \right) & \left( b - c \right)\left( b^2 + bc + a^2 \right)\\1 & c^2 + ab & c^3 \end{vmatrix}\] 
\[= \left( a - b \right)\left( b - c \right)\begin{vmatrix} 0 & a + b - c & a^2 + ab + b^2 \\0 & \left( b + c - a \right) & \left( b^2 + bc + c^2 \right)\\1 & c^2 + ab & c^3 \end{vmatrix} \left[\text{ Taking out }\left( a - b \right)\text{ common from }R_1\text{ and }\left( b - c \right)\text{ from }R_2 \right]\] 
\[ = \left( a - b \right)\left( b - c \right)\begin{vmatrix} 0 & a + b - c & a^2 + ab + b^2 \\0 & \left( b + c - a \right) - \left( a + b - c \right) & \left( b^2 + bc + c^2 \right) - \left( a^2 + ab + b^2 \right)\\1 & c^2 + ab & c^3 \end{vmatrix} \left[\text{ Applying }R \hspace{0.167em}_2 \to R_2 \hspace{0.167em} - R_1 \right]\]
\[= \left( a - b \right)\left( b - c \right)\begin{vmatrix} 0 & a + b - c & a^2 + ab + b^2 \\0 & 2 \left( c - a \right) & b\left( c - a \right) + \left( c^2 - a^2 \right)\\1 & c^2 + ab & c^3 \end{vmatrix}\] 
\[ = \left( a - b \right)\left( b - c \right)\left( c - a \right) \begin{vmatrix}0 & a + b - c & a^2 + ab + b^2 \\0 & 2 & a + b + c\\1 & c^2 + ab & c^3 \end{vmatrix}\] 
\[ = \left( a - b \right)\left( b - c \right)\left( c - a \right) \times \left\{ 1 \times \begin{vmatrix} a + b - c & a^2 + ab + b^2 \\ 2 & a + b + c \end{vmatrix} \right\} \left[\text{ Expanding along }C_1 \right]\]
\[= \left( a - b \right)\left( b - c \right)\left( c - a \right) \times \left\{ \left( a + b \right)^2 - c^2 - \left( 2 a^2 + 2ab + 2 b^2 \right) \right\}\] 
\[ = \left( a - b \right)\left( b - c \right)\left( c - a \right)\left\{ \left( a + b \right)^2 - c^2 - \left( a + b \right)^2 - \left( a^2 + b^2 \right) \right\}\] 
\[ = - \left( a - b \right)\left( b - c \right)\left( c - a \right)\left( a^2 + b^2 + c^2 \right)\] 
\[ = RHS\]
Hence proved.