Prove the Identity : \[ \frac{\sin^3 x+\cos^3 x}{\sin x+\cos x} + \frac{\sin^3 x-\cos^3 x}{\sin x-\cos x} =2 \]

Solution:

\[ \frac{(\sin x+\cos x)(\sin^2 x-\sin x\cos x+\cos^2 x)} {\sin x+\cos x} \]

\[ + \frac{(\sin x-\cos x)(\sin^2 x+\sin x\cos x+\cos^2 x)} {\sin x-\cos x} \]

\[ = \sin^2 x-\sin x\cos x+\cos^2 x \]

\[ + \sin^2 x+\sin x\cos x+\cos^2 x \]

\[ = 2\sin^2 x+2\cos^2 x \]

\[ = 2(\sin^2 x+\cos^2 x) \]

\[ =2 \]

Hence proved.