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(D) Let the given determinant be $\Delta$. Using the identity $(a+b)^2 - (a-b)^2 = 4ab$,we observe that for any term of the form $(n^x + n^{-x})^2 - (

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$0$

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(D) Let the given determinant be $\Delta$. Using the identity $(a+b)^2 - (a-b)^2 = 4ab$,we observe that for any term of the form $(n^x + n^{-x})^2 - (n^x - n^{-x})^2$,the result is $4(n^x)(n^{-x}) = 4(n^0) = 4$. Applying the column operation $C_1 ightarrow C_1 - C_2$,the first column becomes: $C_1 = \begin{bmatrix} (5^x + 5^{-x})^2 - (5^x - 5^{-x})^2 \ (6^x + 6^{-x})^2 - (6^x - 6^{-x})^2 \ (7^x + 7^{-x})^2 - (7^x - 7^{-x})^2 \end{bmatrix} = \begin{bmatrix} 4 \ 4 \ 4 \end{bmatrix}$. Now the determinant is $\left|\begin{array}{ccc} 4 & (5^x - 5^{-x})^2 & 1 \ 4 & (6^x - 6^{-x})^2 & 1 \ 4 & (7^x - 7^{-x})^2 & 1 \end{array}ight|$. Since the first column consists of identical elements $(4)$,we can factor out $4$ from the determinant: $4 \left|\begin{array}{ccc} 1 & (5^x - 5^{-x})^2 & 1 \ 1 & (6^x - 6^{-x})^2 & 1 \ 1 & (7^x - 7^{-x})^2 & 1 \end{array}ight|$. Since column $1$ and column $3$ are identical,the value of the determinant is $4 	imes 0 = 0$.

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