If p = 124, then the value of $$\root 3 \of {p\left( {{p^2} + 3p + 3} \right) + 1} = ?$$

Solution (By Examveda Team)

$$\eqalign{ & p = 124 \cr & \root 3 \of {p\left( {{p^2} + 3p + 3} \right) + 1} \cr & = \root 3 \of {{p^3} + 3{p^2} + 3p + 1} \cr & = \root 3 \of {{{\left( {p + 1} \right)}^3}} \cr & = \root 3 \of {{{\left( {125} \right)}^3}} \cr & = 125 \cr} $$