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(A) Let $\Delta = \left|\begin{array}{ccc}\cos 2x & \sin^2 x & \cos 2x \ \sin^2 x & \cos 2x & \cos^2 x \ \cos 2x & \cos^2 x & \cos 2x\end{array}ig

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(A) Let $\Delta = \left|\begin{array}{ccc}\cos 2x & \sin^2 x & \cos 2x \ \sin^2 x & \cos 2x & \cos^2 x \ \cos 2x & \cos^2 x & \cos 2x\end{array}ight|$.We know that $\cos 2x = 2\cos^2 x - 1$ and $\sin^2 x = 1 - \cos^2 x$.Substituting these into the determinant:$\Delta = \left|\begin{array}{ccc} 2\cos^2 x - 1 & 1 - \cos^2 x & 2\cos^2 x - 1 \ 1 - \cos^2 x & 2\cos^2 x - 1 & \cos^2 x \ 2\cos^2 x - 1 & \cos^2 x & 2\cos^2 x - 1 \end{array}ight|$.Let $u = \cos^2 x$. Then the determinant becomes:$\Delta = \left|\begin{array}{ccc} 2u-1 & 1-u & 2u-1 \ 1-u & 2u-1 & u \ 2u-1 & u & 2u-1 \end{array}ight|$.Subtract $C_3$ from $C_1$:$\Delta = \left|\begin{array}{ccc} 0 & 1-u & 2u-1 \ 1-u-u & 2u-1 & u \ 0 & u & 2u-1 \end{array}ight| = \left|\begin{array}{ccc} 0 & 1-u & 2u-1 \ 1-2u & 2u-1 & u \ 0 & u & 2u-1 \end{array}ight|$.Expanding along $C_1$:$\Delta = -(1-2u) \left[ (1-u)(2u-1) - u(2u-1) ight] = -(1-2u)(2u-1)(1-u-u) = (2u-1)^2(1-2u) = -(2u-1)^3$.Substituting $u = \cos^2 x$:$\Delta = -(2\cos^2 x - 1)^3 = -(8\cos^6 x - 12\cos^4 x + 6\cos^2 x - 1) = -8\cos^6 x + 12\cos^4 x - 6\cos^2 x + 1$.The constant term is the term independent of $\cos x$,which is $1$.

If the determinant $\left|\begin{array}{ccc}\cos 2x & \sin^2 x & \cos 2x \\ \sin^2 x & \cos 2x & \cos^2 x \\ \cos 2x & \cos^2 x & \cos 2x\end{array}\right|$ is expanded in powers of $\cos x$,then the constant term in the expansion is

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