If a, b, c 0 and x, y, z in R,then the valu | vedclass

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(D) Let the given determinant be $\Delta$. We know that $(p+q)^2 - (p-q)^2 = 4pq$. Applying the column operation $C_1 \rightarrow C_1 - C_2$,the first

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$0$

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(D) Let the given determinant be $\Delta$. We know that $(p+q)^2 - (p-q)^2 = 4pq$. Applying the column operation $C_1 ightarrow C_1 - C_2$,the first column becomes: $(a^x + a^{-x})^2 - (a^x - a^{-x})^2 = 4(a^x)(a^{-x}) = 4(a^0) = 4$. Similarly,for the second and third rows,the first column elements become $4$. The determinant now becomes: $\Delta = \left| \begin{array}{ccc} 4 & (a^x - a^{-x})^2 & 1 \ 4 & (b^y - b^{-y})^2 & 1 \ 4 & (c^z - c^{-z})^2 & 1 \end{array} ight|$. Taking $4$ as a common factor from the first column,we get: $\Delta = 4 \left| \begin{array}{ccc} 1 & (a^x - a^{-x})^2 & 1 \ 1 & (b^y - b^{-y})^2 & 1 \ 1 & (c^z - c^{-z})^2 & 1 \end{array} ight|$. Since the first column and the third column are identical,the value of the determinant is $0$.

If $a, b, c > 0$ and $x, y, z \in R$,then the value of the determinant $\left| \begin{array}{ccc} (a^x + a^{-x})^2 & (a^x - a^{-x})^2 & 1 \\ (b^y + b^{-y})^2 & (b^y - b^{-y})^2 & 1 \\ (c^z + c^{-z})^2 & (c^z - c^{-z})^2 & 1 \end{array} \right|$ is equal to:

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