If p = 99, then the value of p(p² + 3p + 3) is?

$$\eqalign{ & \because p = 99 \cr & p\left( {{p^2} + 3p + 3} \right) \cr & = {p^3} + 3{p^2} + 3p + 1 - 1 \cr & = {\left( {p + 1} \right)^3} - 1 \cr & = {\left( {99 + 1} \right)^3} - 1 \cr & = {\left( {100} \right)^3} - 1 \cr & = 1000000 - 1 \cr & = 999999 \cr} $$