If p = 124, then the value of 3 p ( p^2 + 3p + 3 ) + 1 = ?

Answer: Option D

$$\begin{array}{rl}& p=124\\ & \text{root}3\text{of}p\left(p^2+3p+3\right)+1\\ & =\text{root}3\text{of}{p}^3+3p^2+3p+1\\ & =\text{root}3\text{of}{\left(p+1\right)}^3\\ & =\text{root}3\text{of}{\left(125\right)}^3\\ & =125\end{array}$$