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If p = 999, then the value of $$\root 3 \of {p\left( {{p^2} + 3p + 3} \right) + 1} {\text{ is?}}$$
Answer & Solution
Correct Answer:
Option
A
$$\eqalign{
& \because p = 999 \cr
& \root 3 \of {p\left( {{p^2} + 3p + 3} \right) + 1} \cr
& = \root 3 \of {{p^3} + 3{p^2} + 3p + 1} \cr
& = \root 3 \of {{{\left( {p + 1} \right)}^3}} \cr
& = \root 3 \of {{{\left( {999 + 1} \right)}^3}} \cr
& = \root 3 \of {{{\left( {1000} \right)}^3}} \cr
& = 1000 \cr} $$
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