Without expanding, show that the value of the determinant is zero:
`|[(2^x+2^(-x))^2, (2^x-2^(-x))^2, 1] , [(3^x+3^(-x))^2, (3^x-3^(-x))^2, 1] , [(4^x+4^(-x))^2, (4^x-4^(-x))^2, 1]|`

Explanation

$$\left|\begin{array}{lll}\left(2^{x}+2^{-x}\right)^{2} & \left(2^{x}-2^{-x}\right)^{2} & 1 \\ \left(3^{x}+3^{-x}\right)^{2} & \left(3^{x}-3^{-x}\right)^{2} & 1 \\ \left(4^{x}+4^{-x}\right)^{2} & \left(4^{x}-4^{-x}\right)^{2} & 1\end{array}\right|$$
$$=\left|\begin{array}{lll}\left(2^{2 x}+2^{-2 x}+2\right) & \left(2^{2 x}+2^{-2 x}-2\right) & 1 \\ \left(3^{2 x}+3^{-2 x}+2\right) & \left(3^{2 x}+3^{-2 x}-2\right) & 1 \\ \left(4^{2 x}+4^{-2 x}+2\right) & \left(4^{2 x}+4^{-2 x}-2\right) & 1 \\ 4 & \left(2^{2 x}+2^{-2 x}-2\right) & 1\end{array}\right|$$
$$=\left|\begin{array}{lll}4 & \left(2^{2 x}+2^{-2 x}-2\right) & 1 \\ 4 & \left(3^{2 x}+3^{-2 x}-2\right) & 1 \\ 4 & \left(4^{2 x}+4^{-2 x}-2\right) & 1\end{array}\right|\left[\right.$$ Applying $$\left.C_{1} \rightarrow C_{1}-C_{2}\right]$$
$$=4\left|\begin{array}{lll}1 & \left(2^{2 x}+2^{-2 x}-2\right) & 1 \\ 1 & \left(3^{2 x}+3^{-2 x}-2\right) & 1 \\ 1 & \left(4^{2 x}+4^{-2 x}-2\right) & 1\end{array}\right|$$
$$=0$$