Question

For $p=99$ then find the value of $p^3+3p^2+3p$

Answer 1

We have,

$p=99$

Since,

$\Rightarrow p^3+3p^2+3p+1-1$

$\Rightarrow {(p+1)}^3-1$

$\Rightarrow {(99+1)}^3-1$

$\Rightarrow {\left(100\right)}^3-1$

$\Rightarrow 1000000-1$

$\Rightarrow 999999$

Hence, this is the answer.