Step 1:
Let $\large\frac{\sin^3x}{1+\cos x}+\frac{\cos^3x}{1-\sin x}$$=A$
Then
$A=\large\frac{(\sin^3x+\cos^3x)+(\cos^4x-\sin^4x)}{(1+\cos x)(1-\sin x)}$
$A=\large\frac{(\sin^3x+\cos^3x)\big((\cos x+\sin x)(\cos x-\sin x)(\cos^2x+\sin^2x)\big)}{(1+\cos x)(1-\sin x)}$
$A=\large\frac{(\sin x+\cos x)\{(1-\sin x\cos x)+(\cos x-\sin x)}{1+\cos x-\sin x-\sin x\cos x}$
Step 2:
$A=\sin x+\cos x$
$A=\sqrt 2\big[\large\frac{1}{\sqrt 2}$$\sin x+\large\frac{1}{\sqrt 2}$$\cos x\big]$------(1)
$A=\sqrt 2\big[\cos\large\frac{\pi}{4}$$\sin x+\sin\large\frac{\pi}{4}$$\cos x\big]$
$\Rightarrow \sqrt 2\sin\big[\large\frac{\pi}{4}$$+x]$
Step 3:
By equation (1) we get
$A=\sqrt 2\big[\sin\large\frac{\pi}{4}$$\sin x+\cos\large\frac{\pi}{4}$$\cos x\big]$
$\;\;=\sqrt 2\cos\big[\large\frac{\pi}{4}$$-x\big]$
Hence (a) is the correct option.