If $\Delta = \begin{vmatrix} 1 & a & a^2 \\ a & a^2 & 1 \\ a^2 & 1 & a \end{vmatrix} = -4$, then find the value of $\begin{vmatrix} a^3 - 1 & 0 & a-a^4 \\ 0 & a-a^4 & a^3-1 \\ a-a^4 & a^3-1 & 0 \end{vmatrix}$
$\left[\text{Note: If } C_{ij} \text{ are the cofactors of } a_{ij} \text{ in } \left|A\right| = \begin{vmatrix}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{vmatrix} \\
\text{ then }
\begin{vmatrix}
C_{11} & C_{12} & C_{13} \\
C_{21} & C_{22} & C_23 \\
C_{31} & C_{32} & C_{33}
\end{vmatrix} = \left|A\right|^2\right] $
Cofactors of determinant $\Delta$ are
$
C_{11} = \left(-1\right)^{1+1}
\begin{vmatrix}
a^2 & 1 \\
1 & a
\end{vmatrix} = a^3 - 1 $
$
C_{12} = \left(-1\right)^{1+2}
\begin{vmatrix}
a & 1 \\
a^2 & a
\end{vmatrix} = a^2-a^2=0$
$C_{13} = \left(-1\right)^{1+3} \begin{vmatrix}
a & a^2 \\
a^2 & 1
\end{vmatrix} = a - a^4$
$C_{21} = \left(-1\right)^{2+1} \begin{vmatrix}
a & a^2 \\
1 & a
\end{vmatrix} = a^2 - a^2 = 0$
$C_{22} = \left(-1\right)^{2+2} \begin{vmatrix}
1 & a^2 \\
a^2 & a
\end{vmatrix} = a - a^4$
$C_{23} = \left(-1\right)^{2+3} \begin{vmatrix}
1 & a \\
a^2 & 1
\end{vmatrix} = -\left(1-a^3\right) = a^3 - 1$
$C_{31} = \left(-1\right)^{3+1} \begin{vmatrix}
a & a^2 \\
a^2 & 1
\end{vmatrix} = a - a^4$
$C_{32} = \left(-1\right)^{3+2} \begin{vmatrix}
1 & a^2 \\
a & 1
\end{vmatrix} = -\left(1-a^3\right) = a^3 - 1$
$C_{33} = \left(-1\right)^{3+3} \begin{vmatrix}
1 & a \\
a & a^2
\end{vmatrix} = a^2 - a^2 = 0$
Now, $\begin{vmatrix}
a^3 - 1 & 0 & a-a^4 \\
0 & a-a^4 & a^3-1 \\
a-a^4 & a^3-1 & 0
\end{vmatrix} = \begin{vmatrix}
C_{11} & C_{12} & C_{13} \\
C_{21} & C_{22} & C_{23} \\
C_{31} & C_{32} & C_{33}
\end{vmatrix}$
$\therefore$
$\begin{vmatrix}
a^3 - 1 & 0 & a-a^4 \\
0 & a-a^4 & a^3-1 \\
a-a^4 & a^3-1 & 0
\end{vmatrix} = \left|\Delta\right|^2 = \left(-4\right)^2 = 16$