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

19999
2999999
399999
4999

Answer:- 1

Explanation:-

Solution:

\begin{aligned} ∵ p = 99 \\ p (p^{2} + 3 p + 3) \\ = p^{3} + 3 p^{2} + 3 p + 1 - 1 \\ = {(p + 1)}^{3} - 1 \\ = {(99 + 1)}^{3} - 1 \\ = {(100)}^{3} - 1 \\ = 1000000 - 1 \\ = 999999 \end{aligned}

$\eqalign{ & \because p = 99 \cr & p\left( {{p^2} + 3p + 3} \right) \cr & = {p^3} + 3{p^2} + 3p + 1 - 1 \cr & = {\left( {p + 1} \right)^3} - 1 \cr & = {\left( {99 + 1} \right)^3} - 1 \cr & = {\left( {100} \right)^3} - 1 \cr & = 1000000 - 1 \cr & = 999999 \cr}$

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", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} ∵ p = 99 \\ p (p^{2} + 3 p + 3) \\ = p^{3} + 3 p^{2} + 3 p + 1 - 1 \\ = {(p + 1)}^{3} - 1 \\ = {(99 + 1)}^{3} - 1 \\ = {(100)}^{3} - 1 \\ = 1000000 - 1 \\ = 999999 \end{aligned}

$\eqalign{ & \because p = 99 \cr & p\left( {{p^2} + 3p + 3} \right) \cr & = {p^3} + 3{p^2} + 3p + 1 - 1 \cr & = {\left( {p + 1} \right)^3} - 1 \cr & = {\left( {99 + 1} \right)^3} - 1 \cr & = {\left( {100} \right)^3} - 1 \cr & = 1000000 - 1 \cr & = 999999 \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM" } }

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