Question

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

Answer 1

\[∆ = \begin{vmatrix}- a( b^2 + c^2 - a^2 ) & 2 b^3 & 2 c^3 \\ 2 a^3 & - b( c^2 + a^2 - b^2 ) & 2 c^3 \\ 2 a^3 & 2 b^3 & - c( a^2 + b^2 - c^2 )\end{vmatrix}\]

\[ = abc\begin{vmatrix}- b^2 - c^2 + a^2 & 2 b^2 & 2 c^2 \\ 2 a^2 & - c^2 - a^2 + b^2 & 2 c^2 \\ 2 a^2 & 2 b^2 & - a^2 - b^2 + c^2\end{vmatrix} \left[\text{ Taking out a, b and c common from }C_1 , C_2\text{ and }|C_3 \right]\]

\[ = abc\begin{vmatrix}a^2 + b^2 + c^2 & 2 b^2 & 2 c^2 \\ a^2 + b^2 + c^2 & - c^2 - a^2 + b^2 & 2 c^2 \\ a^2 + b^2 + c^2 & 2 b^2 & - a^2 - b^2 + c^2\end{vmatrix} \left[\text{ Applying }C_1\text{ to }C_1 + C_2 + C_3 \right]\]

\[ = abc( a^2 + b^2 + c^2 )\begin{vmatrix}1 & 2 b^2 & 2 c^2 \\ 1 & - c^2 - a^2 + b^2 & 2 c^2 \\ 1 & 2 b^2 & - a^2 - b^2 + c^2\end{vmatrix} \left[\text{ Taking out }a^2 + b^2 + c \text{ common from }C_1 \right]\]

\[ = abc( a^2 + b^2 + c^2 )\begin{vmatrix}1 & 2 b^2 & 2 c^2 \\ 0 & - c^2 - a^2 - b^2 & 0 \\ 0 & 0 & - a^2 - b^2 - c^2\end{vmatrix} \left[\text{ Applying }R_2 \text{ to }R_2 - R_1 \text{ and }R_3 \text{ to }R_3 - R_1 \right]\]

\[ = abc( a^2 + b^2 + c^2 )^3 \left[\text{ Expanding }\right]\]

Hence proved.