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

1999999
21000000
31000001
4999998

Answer:- 1

Explanation:-

Solution:

\begin{aligned} p (p^{2} + 3 p + 3) \\ = p^{3} + 3 p^{2} + 3 p + 1 - 1 \end{aligned}

$\eqalign{ & p\left( {{p^2} + 3p + 3} \right) \cr & = {p^3} + 3{p^2} + 3p + 1 - 1 \cr}$
(On adding and subtracting one both side)

\begin{aligned} = {(p + 1)}^{3} - 1 \\ = {(99 + 1)}^{3} - 1 \\ = 100^{3} - 1 \\ = 1000000 - 3 \\ = 99999 \end{aligned}

$\eqalign{ & = {\left( {p + 1} \right)^3} - 1 \cr & = {\left( {99 + 1} \right)^3} - 1 \cr & = {100^3} - 1 \cr & = 1000000 - 3 \cr & = 99999 \cr}$

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(On adding and subtracting one both side)

\begin{aligned} = {(p + 1)}^{3} - 1 \\ = {(99 + 1)}^{3} - 1 \\ = 100^{3} - 1 \\ = 1000000 - 3 \\ = 99999 \end{aligned}

$\eqalign{ & = {\left( {p + 1} \right)^3} - 1 \cr & = {\left( {99 + 1} \right)^3} - 1 \cr & = {100^3} - 1 \cr & = 1000000 - 3 \cr & = 99999 \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} p (p^{2} + 3 p + 3) \\ = p^{3} + 3 p^{2} + 3 p + 1 - 1 \end{aligned}

$\eqalign{ & p\left( {{p^2} + 3p + 3} \right) \cr & = {p^3} + 3{p^2} + 3p + 1 - 1 \cr}$
(On adding and subtracting one both side)

\begin{aligned} = {(p + 1)}^{3} - 1 \\ = {(99 + 1)}^{3} - 1 \\ = 100^{3} - 1 \\ = 1000000 - 3 \\ = 99999 \end{aligned}

$\eqalign{ & = {\left( {p + 1} \right)^3} - 1 \cr & = {\left( {99 + 1} \right)^3} - 1 \cr & = {100^3} - 1 \cr & = 1000000 - 3 \cr & = 99999 \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM" } }