If \(\sin\theta + \cos\theta = \sqrt{3}\), then the value of \(\sin^{3}\theta + \cos^{3}\theta\) is
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Use the identity \(a^{3} + b^{3} = (a + b)(a^{2} - ab + b^{2})\) with \(a = \sin\theta\) and \(b = \cos\theta\).
Square the given relation: \((\sin\theta + \cos\theta)^{2} = 3\), so \(\sin^{2}\theta + 2\sin\theta\cos\theta + \cos^{2}\theta = 3\), giving \(1 + 2\sin\theta\cos\theta = 3\) and \(\sin\theta\cos\theta = 1\).
Now \(\sin^{3}\theta + \cos^{3}\theta = (\sin\theta + \cos\theta)(\sin^{2}\theta - \sin\theta\cos\theta + \cos^{2}\theta) = \sqrt{3}(1 - 1) = 0\).
Hence, \(\sin^{3}\theta + \cos^{3}\theta = 0\).
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