# How do you factor Sin3x - Cos3x?

We have to factorise sin3x – cos3x

### Solution

$\sin ^{3}x -\cos ^{3}x$

Let us solve using the identity

(a3 – b3) = (a – b)(a2 + b2 + ab)

Here a =sin x and b = cos x

So

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

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

Substituting the identity in above equation we get,

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

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