Question

\[\begin{vmatrix}1 & a & a^2 \\ a^2 & 1 & a \\ a & a^2 & 1\end{vmatrix} = \left( a^3 - 1 \right)^2\]

Answer 1

\[\text{ Let LHS }= \Delta = \begin{vmatrix} 1 & a & a^2 \\ a^2 & 1 & a\\a & a^2 & 1 \end{vmatrix}\] 
\[\Delta = \begin{vmatrix} 1 + a^2 + a & 1 + a^2 + a & 1 + a^2 + a\\ a^2 & 1 & a\\a & a^2 & 1 \end{vmatrix} \left[\text{ Applyng }R_1 \to R_1 + R_2 + R_2 \right]\] 
\[ = \left( 1 + a^2 + a \right) \begin{vmatrix} 1 & 1 & 1 \\ a^2 & 1 & a\\a & a^2 & 1 \end{vmatrix} \left[\text{ Applying }C_2 \to C_2 - C_1\text{ and }C_3 \to C_3 - C_1 \right]\] 
\[ = \left( 1 + a^2 + a \right) \begin{vmatrix} 1 & 0 & 0 \\ a^2 & 1 - a^2 & a - a^2 \\a & a^2 - a & 1 - a \end{vmatrix}\] 
\[ = \left( 1 + a^2 + a \right) \begin{vmatrix} 1 & 0 & 0 \\ a^2 & \left( 1 - a \right)\left( 1 + a \right) & a\left( 1 - a \right)\\a & a\left( a - 1 \right) & 1 - a \end{vmatrix}\] 
\[ = \left( 1 + a^2 + a \right)\left( a - 1 \right)\left( a - 1 \right) \begin{vmatrix} 1 & 0 & 0\\ a^2 & - \left( 1 + a \right) & - a\\a & a & - 1 \end{vmatrix} \left[\text{ Taking out (a - 1) common from }C_2\text{ and }C_3 \right]\]
\[ = \left( a^3 - 1 \right)\left\{ \left( a - 1 \right) \begin{vmatrix} 1 & 0 & 0\\a & - \left( 1 + a \right) & - a\\a & a & - 1 \end{vmatrix} \right\} \left[ \because \left( 1 + a^2 + a \right)\left( a - 1 \right) = \left( a^3 - 1 \right) \right]\]
\[ = \left( a^3 - 1 \right)\left\{ \left( a - 1 \right)\left( 1 + a^{} + a^2 \right) \right\}\] 
\[ = \left( a^3 - 1 \right)\left( a^3 - 1 \right)\] 
\[ = \left( a^3 - 1 \right)^2 \] 
\[ = RHS \] 

Hence proved.