If $x, y, z \in R$, the value of the determinant$\left|\begin{array}{lll}\left(2^{x}+2^{-x}\right)^{2} & \left(2^{x}-2^{-x}\right)^{2} & 1 \\ \left(3^{x}+3^{-x}\right)^{2} & \left(3^{x}-3^{-x}\right)^{2} & 1 \\ \left(4^{x}+4^{-x}\right)^{2} & \left(4^{x}-4^{-x}\right)^{2} & 1\end{array}\right|$is equal to_____________
Let $\Delta=\left|\begin{array}{lll}\left(2^{x}+2^{-x}\right)^{2} & \left(2^{x}-2^{-x}\right)^{2} & 1 \\ \left(3^{x}+3^{-x}\right)^{2} & \left(3^{x}-3^{-x}\right)^{2} & 1 \\ \left(4^{x}+4^{-x}\right)^{2} & \left(4^{x}-4^{-x}\right)^{2} & 1\end{array}\right|$
$\Delta=\left|\begin{array}{lll}\left(2^{x}+2^{-x}\right)^{2} & \left(2^{x}-2^{-x}\right)^{2} & 1 \\ \left(3^{x}+3^{-x}\right)^{2} & \left(3^{x}-3^{-x}\right)^{2} & 1 \\ \left(4^{x}+4^{-x}\right)^{2} & \left(4^{x}-4^{-x}\right)^{2} & 1\end{array}\right|$
Applying $C_{1} \rightarrow C_{1}-C_{2}$
$=\left|\begin{array}{lll}\left(2^{x}+2^{-x}\right)^{2}-\left(2^{x}-2^{-x}\right)^{2} & \left(2^{x}-2^{-x}\right)^{2} & 1 \\ \left(3^{x}+3^{-x}\right)^{2}-\left(3^{x}-3^{-x}\right)^{2} & \left(3^{x}-3^{-x}\right)^{2} & 1 \\ \left(4^{x}+4^{-x}\right)^{2}-\left(4^{x}-4^{-x}\right)^{2} & \left(4^{x}-4^{-x}\right)^{2} & 1\end{array}\right|$
$=\left|\begin{array}{lll}\left(2^{x}+2^{-x}+2^{x}-2^{-x}\right)\left(2^{x}+2^{-x}-2^{x}+2^{-x}\right) & \left(2^{x}-2^{-x}\right)^{2} & 1 \\ \left(3^{x}+3^{-x}+3^{x}-3^{-x}\right)\left(3^{x}+3^{-x}-3^{x}+3^{-x}\right) & \left(3^{x}-3^{-x}\right)^{2} & 1 \\ \left(4^{x}+4^{-x}+4^{x}-4^{-x}\right)\left(4^{x}+4^{-x}-4^{x}+4^{-x}\right) & \left(4^{x}-4^{-x}\right)^{2} & 1\end{array}\right|$
$=\left|\begin{array}{lll}\left(2^{x}+2^{x}\right)\left(2^{-x}+2^{-x}\right) & \left(2^{x}-2^{-x}\right)^{2} & 1 \\ \left(3^{x}+3^{x}\right)\left(3^{-x}+3^{-x}\right) & \left(3^{x}-3^{-x}\right)^{2} & 1 \\ \left(4^{x}+4^{x}\right)\left(4^{-x}+4^{-x}\right) & \left(4^{x}-4^{-x}\right)^{2} & 1\end{array}\right|$
$=\left|\begin{array}{lll}\left(2.2^{x}\right)\left(2.2^{-x}\right) & \left(2^{x}-2^{-x}\right)^{2} & 1 \\ \left(2.3^{x}\right)\left(2.3^{-x}\right) & \left(3^{x}-3^{-x}\right)^{2} & 1 \\ \left(2.4^{x}\right)\left(2.4^{-x}\right) & \left(4^{x}-4^{-x}\right)^{2} & 1\end{array}\right|$
Taking out $(4)$ common from $C_{1}$
$=4\left|\begin{array}{lll}\left(2^{x}\right)\left(2^{-x}\right) & \left(2^{x}-2^{-x}\right)^{2} & 1 \\ \left(3^{x}\right)\left(3^{-x}\right) & \left(3^{x}-3^{-x}\right)^{2} & 1 \\ \left(4^{x}\right)\left(4^{-x}\right) & \left(4^{x}-4^{-x}\right)^{2} & 1\end{array}\right|$
$=4\left|\begin{array}{lll}1 & \left(2^{x}-2^{-x}\right)^{2} & 1 \\ 1 & \left(3^{x}-3^{-x}\right)^{2} & 1 \\ 1 & \left(4^{x}-4^{-x}\right)^{2} & 1\end{array}\right|$
$=4(0)$ ( $\because$ if two columns are identical then the value of determinant is zero)
$=0$
Hence, the value of the determinant $\left|\begin{array}{lll}\left(2^{x}+2^{-x}\right)^{2} & \left(2^{x}-2^{-x}\right)^{2} & 1 \\ \left(3^{x}+3^{-x}\right)^{2} & \left(3^{x}-3^{-x}\right)^{2} & 1 \\ \left(4^{x}+4^{-x}\right)^{2} & \left(4^{x}-4^{-x}\right)^{2} & 1\end{array}\right|$ is equal to $\underline{0}$.