If p = 999, then the value of 3 p ( p^2 + 3p + 3 ) + 1 is?

Answer: Option A

$$\begin{array}{rl}& \because p=999\\ & \text{root}3\text{of}p\left(p^2+3p+3\right)+1\\ & =\text{root}3\text{of}{p}^3+3p^2+3p+1\\ & =\text{root}3\text{of}{\left(p+1\right)}^3\\ & =\text{root}3\text{of}{\left(999+1\right)}^3\\ & =\text{root}3\text{of}{\left(1000\right)}^3\\ & =1000\end{array}$$