The value of 4x^3 - x ( 2x + 1 ) ( 6x - 3 ) when x = 9999 is?

Answer: Option C

$$\begin{array}{rl}& x=9999\left(\text{ Given}\right)\\ & \frac{4x^3-x}{\left(2x+1\right)\left(6x-3\right)}\\ & =\frac{x\left(4x^2-1\right)}{3\left(2x+1\right)\left(2x-1\right)}\\ & =\frac{x\left(4x^2-1\right)}{3\left(4x^2-1\right)}\\ & =\frac{x}{3}\\ & \therefore \frac{9999}{3}=3333\end{array}$$