If $p = 99$ then find the value of $p^{3} + 3 p^{2} + 3 p$ .

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Solution

Now,
$p^{3} + 3 p^{2} + 3 p$
$= p^{3} + 3 p^{2} + 3 p + 1 - 1$
$= (p + 1)^{3} - 1$
$= 100^{3} - 1$ [ Since $p = 99$ ]
$= 999, 999$ .

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