If x, y in R and left|begin{array}{lll}left(a | vedclass

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(A) Let the given determinant be $D$. Using the identity $(p+q)^2 - (p-q)^2 = 4pq$,we apply the column operation $C_1 	o C_1 - C_2$: $D = \left|\begi

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-$3$

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(A) Let the given determinant be $D$. Using the identity $(p+q)^2 - (p-q)^2 = 4pq$,we apply the column operation $C_1 	o C_1 - C_2$: $D = \left|\begin{array}{lll} (a^x+a^{-x})^2 - (a^x-a^{-x})^2 & (a^x-a^{-x})^2 & 1 \ (b^x+b^{-x})^2 - (b^x-b^{-x})^2 & (b^x-b^{-x})^2 & 1 \ (c^x+c^{-x})^2 - (c^x-c^{-x})^2 & (c^x-c^{-x})^2 & 1 \end{array}ight|$ $D = \left|\begin{array}{lll} 4(a^x)(a^{-x}) & (a^x-a^{-x})^2 & 1 \ 4(b^x)(b^{-x}) & (b^x-b^{-x})^2 & 1 \ 4(c^x)(c^{-x}) & (c^x-c^{-x})^2 & 1 \end{array}ight| = \left|\begin{array}{lll} 4 & (a^x-a^{-x})^2 & 1 \ 4 & (b^x-b^{-x})^2 & 1 \ 4 & (c^x-c^{-x})^2 & 1 \end{array}ight|$ Since two columns ($C_1$ and $C_3$) are proportional (specifically,$C_1 = 4C_3$),the value of the determinant $D = 0$. Given $D = 2y + 6$,we have $0 = 2y + 6$. $2y = -6 \implies y = -3$.

If $x, y \in R$ and $\left|\begin{array}{lll}\left(a^x+a^{-x}\right)^2 & \left(a^x-a^{-x}\right)^2 & 1 \\ \left(b^x+b^{-x}\right)^2 & \left(b^x-b^{-x}\right)^2 & 1 \\ \left(c^x+c^{-x}\right)^2 & \left(c^x-c^{-x}\right)^2 & 1\end{array}\right| = 2y+6$,then $y=$

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