Question:
Prove the following identities (1-16)
$\frac{\sin ^{3} x+\cos ^{3} x}{\sin x+\cos x}+\frac{\sin ^{3} x-\cos ^{3} x}{\sin x-\cos x}=2$
Solution:
$\mathrm{LHS}=\frac{\sin ^{3} x+\cos ^{3} x}{\sin x+\cos x}+\frac{\sin ^{3} x-\cos ^{3} x}{\sin x-\cos x}$
$=\frac{(\sin x+\cos x)\left[\sin ^{2} x+\cos ^{2} x-\sin x \cos x\right]}{(\sin x+\cos x)}+\frac{(\sin x-\cos x)\left[\sin ^{2} x+\cos ^{2} x+\sin x \cos x\right]}{(\sin x-\cos x)}$
$=\left[\sin ^{2} x+\cos ^{2} x-\sin x \cos x\right]+\left[\sin ^{2} x+\cos ^{2} x+\sin x \cos x\right]$
$=[1-\sin x \cos x]+[1+\sin x \cos x]$
$=2$
$=\mathrm{RHS}$
Hence proved.
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