If \sin \theta + \cos \theta = \sqrt{2}, then what is \sin^{6}\theta + \cos^{6}\theta + 6 \sin^{2}\theta \cos^{2}\theta equal to?

Explanation

Squaring \sin \theta + \cos \theta = \sqrt{2} yields 1 + 2\sin\theta\cos\theta = 2, so \sin\theta\cos\theta = \frac{1}{2}. The expression can be rewritten using the identity a^3+b^3 = (a+b)(a^2-ab+b^2). Here, (\sin^2\theta)^3 + (\cos^2\theta)^3 = (\sin^2\theta+\cos^2\theta)((\sin^2\theta+\cos^2\theta)^2 - 3\sin^2\theta\cos^2\theta) = 1(1 - 3(\frac{1}{4})) = \frac{1}{4}. Adding 6\sin^2\theta\cos^2\theta = 6(\frac{1}{4}) = \frac{6}{4} to it gives \frac{1}{4} + \frac{6}{4} = \frac{7}{4}.

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