Evaluate the determinant: $\left| \begin{array}{ccc} (a^x + a^{-x})^2 & (a^x - a^{-x})^2 & 1 \\ (b^x + b^{-x})^2 & (b^x - b^{-x})^2 & 1 \\ (c^x + c^{-x})^2 & (c^x - c^{-x})^2 & 1 \end{array} \right|$

Statement-$1$: The determinant of a skew-symmetric matrix of order $3$ is zero.
Statement-$2$: For any square matrix $A$ of order $n$,$\det(A^T) = \det(A)$ and $\det(-A) = (-1)^n \det(A)$.

Using properties of determinants,prove that:
$\left|\begin{array}{ccc}1 & 1+p & 1+p+q \\ 2 & 3+2 p & 4+3 p+2 q \\ 3 & 6+3 p & 10+6 p+3 q\end{array}\right|=1$

If $\left|\begin{array}{ccc}a^{2} & b c & c^{2}+a c \\ a^{2}+a b & b^{2} & c a \\ a b & b^{2}+b c & c^{2}\end{array}\right|=k a^{2} b^{2} c^{2}$, then $k=$

$A$ value of $\theta \in (0, \pi /3)$,for which $\left| \begin{array}{ccc} 1 + \cos^2 \theta & \sin^2 \theta & 4 \cos 6\theta \\ \cos^2 \theta & 1 + \sin^2 \theta & 4 \cos 6\theta \\ \cos^2 \theta & \sin^2 \theta & 1 + 4 \cos 6\theta \end{array} \right| = 0$,is

If $A$ and $B$ are two square matrices with $\det(A) = 5$ and $\det(B^T \cdot A^T) = -15$,then $\det(B)$ is equal to