Question

If it's given that p = 99, then find the value of the given expression : p(p² + 3p + 3).

Answer is – 999999

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Answer 1

AnswEr :

☯$\:\:\:\underline{\large{\mathcal{METHOD \:\:\sf1 :}}}$

⠀

$\bigstar\mathfrak{\:\:given}\:\sf :p = 99 \\\\:\implies\tt p( {p}^{2} + 3p + 3)\\\\{\scriptsize\qquad\bf{\dag}\:\textsf{putting the value of p.}}\\\\:\implies\tt 99\lbrack(99)^{2} + 3(99) + 3\rbrack\\\\\\:\implies\tt 99(9801 + 297 + 3)\\\\\\:\implies\tt 99 \times 10101 \\\\\\:\implies\large\underline{\boxed{\textsf{\textbf{999,999}}}}$

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☯$\:\:\:\underline{\large{\mathcal{METHOD \:\:\sf2 :}}}$

⠀

$\bigstar\mathfrak{\:\:given}\:\sf :p = 99 \\\\:\implies\tt p( {p}^{2} + 3p + 3)\\\\\\:\implies\tt p^{3} + 3p^{2} + 3p\\\\{\scriptsize\qquad\bf{\dag}\:\textsf{Adding \& Subtracting 1.}}\\\\:\implies\tt p^{3} + 3p^{2} + 3p + 1 - 1 \\\\\\ :\implies\tt (p^{3} + 3p^{2} + 3p + 1) - 1\\\\\\:\implies\tt \lbrack(p)^{3} + (3 \times p^{2} \times 1) +(3 \times p \times 1) + (1)^{3}\rbrack- 1\\\\{\scriptsize\qquad\bf{\dag}\:\:\sf{(a + b)}^{3} =a^3 + 3a^2b + 3ab^2 +b^3}\\\\:\implies\tt (p + 1)^3 - 1\\\\{\scriptsize\qquad\bf{\dag}\:\textsf{putting the value of p.}}\\\\:\implies\tt (99 + 1)^3 - 1\\\\\\:\implies\tt (100)^3 - 1\\\\\\:\implies\tt 1000000 - 1\\\\\\:\implies\large\underline{\boxed{\textsf{\textbf{999,999}}}}$

Answer 2

Answer:

Value of p = 99

= p(p² + 3p + 3)

So, we substitute the 99 to p,

= 99(99² + 3(99) + 3)

= 99(9801 + 297 + 3)

= 99(10101)

= 999999

∴ Therefore, the answer is 999999