If p = 999, then the value of $$\root 3 \of {p\left( {{p^2} + 3p + 3} \right) + 1} {\text{ is?}}$$

A. 1000

B. 999

C. 998

D. 1002

Answer: Option A

Solution(By Examveda Team)

$$\eqalign{ & \because p = 999 \cr & \root 3 \of {p\left( {{p^2} + 3p + 3} \right) + 1} \cr & = \root 3 \of {{p^3} + 3{p^2} + 3p + 1} \cr & = \root 3 \of {{{\left( {p + 1} \right)}^3}} \cr & = \root 3 \of {{{\left( {999 + 1} \right)}^3}} \cr & = \root 3 \of {{{\left( {1000} \right)}^3}} \cr & = 1000 \cr} $$