What is the value of $$\frac{(32 \cos^6 x - 48 \cos^4 x + 18 \cos^2 x - 1)}{[4 \sin x \cos x \sin (60 - x) \cos (60 - x) \sin (60 + x) \cos (60 + x)]}$$?

We know :

$$\cos\ \left(2X\right)=2\cos^2X-1\ .$$

Now replace X=3x :

$$\cos6X=2\cos^23X-1\ .$$

Again, we know : $$\cos3X=4\cos^3X-3\cos X\ .$$

So, $$\cos6X=2\left(4\cos^3X-3\cos X\ \right)^2-1\ .$$

or, $$\cos6X=2\left(16\cos^6X+9\cos^2X\ -24\cos^3X\ \cos X\right)-1\ .$$

or, $$\cos6X=32\cos^6X+18\cos^2X\ -48\cos^4X\ -1\ .$$

Now, 

$$4 sin (60-x).sin x. sin(60+x)$$

$$= 4 (sin 60.cos x - cos60 sin x).sin x. (sin 60.cos x + cos60 sin x)$$

$$=4\left(\frac{\sqrt{3}}{2}.\cos x-\sin x\times\frac{1}{2}\right).\sin x.\ \left(\frac{\sqrt{3}}{2}.\cos x+\sin x\times\frac{1}{2}\right)$$

$$= 4 sin x.(3/4. cos^2x - sin ^2x\times1/4)$$

$$= sin x.(3 cos^2x - sin ^2x)$$

$$= sin x [cos^2 x - sin^2 x + 2.cos^2 x]$$

$$= sin x.[cos^2 x - sin^2 x] + cos x.2 sin x.cos x$$

$$= sin x. cos 2x + cos x.sin 2x$$

$$= sin (x + 2x)$$

$$= sin 3x$$ 

Similarly, $$4\cos x.\cos\left(60+x\right).\cos\left(60-x\right)=\cos3x\ .$$

So, $$\frac{(32\cos^6x-48\cos^4x+18\cos^2x-1)}{[4\sin x\cos x\sin(60-x)\cos(60-x)\sin(60+x)\cos(60+x)]}$$

$$=\frac{4\times2\times\cos6x}{2\times\sin3x\times\cos3x}$$

$$=8\cot6x\ .$$

C is correct choice.