How do you factor Sin3x - Cos3x?

We have to factorise sin3x – cos3x


\(\sin ^{3}x -\cos ^{3}x\)

Let us solve using the identity

(a3 – b3) = (a – b)(a2 + b2 + ab)

Here a =sin x and b = cos x


\(\sin ^{3}x -\cos ^{3}x=(\sin x -\cos x)(\sin^{2}x + \sin x\cos x +\cos ^{2}x )\)

We know that \(sin^{2}x +\cos ^{2}x =1\)

Substituting the identity in above equation we get,

\(\sin ^{3}x -\cos ^{3}x=(\sin x -\cos x)( 1 +\sin x\cos x )\)

=\(=\frac{1}{2}(\sin x -\cos x)( 2 +\sin(2x))\) \(\sin ^{3}x -\cos ^{3}x=\frac{1}{2}(\sin x -\cos x)( 2 +\sin(2x))\)

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