Question

Prove the following identities (1-16)
$\frac{{\sin }^{3}x+{\cos }^{3}x}{\sin x+\cos x}+\frac{{\sin }^{3}x-{\cos }^{3}x}{\sin x-\cos x}=2$

Answer 1

$$\begin{gathered} \mathrm{LHS}=\frac{{\sin }^{3}x+{\cos }^{3}x}{\sin x+\cos x}+\frac{{\sin }^{3}x-{\cos }^{3}x}{\sin x-\cos x} \\ =\frac{\left(\sin x+\cos x\right)\left[{\sin }^{2}x+{\cos }^{2}x-\sin x\cos x\right]}{\left(\sin x+\cos x\right)}+\frac{\left(\sin x-\cos x\right)\left[{\sin }^{2}x+{\cos }^{2}x+\sin x\cos x\right]}{\left(\sin x-\cos x\right)} \\ =\left[{\sin }^{2}x+{\cos }^{2}x-\sin x\cos x\right]+\left[{\sin }^{2}x+{\cos }^{2}x+\sin x\cos x\right] \\ =\left[1-\sin x\cos x\right]+\left[1+\sin x\cos x\right] \\ =2 \\ =\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$