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If $$A = (\cos x + \sin x)^2 + (\cos x - \sin x)^2$$, $$B = 2(\cos^6 x + \sin^6 x) - 3(\cos^4 x + \sin^4 x) + 1$$ then
$$A = 2(\cos^2 x + \sin^2 x) = 2$$. With $$s = \sin x$$, $$c = \cos x$$, $$B = 2(1 - 3s^2c^2) - 3(1 - 2s^2c^2) + 1 = 0$$. So $$A \ne B$$, $$A^2 + B = 4$$, $$A^2 - B^2 = 4$$, and $$B^2 - 3B = 0$$ makes the product in d equal to 0. Hence options b, c and d are correct.
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