Question

If P = 999, then $\sqrt[3]{3P(P^2+3P+3)+1}$ =

• A. 1000
• B. 999
• C. 1002
• D. 998

Answer 1

The correct option is A 1000

P = 999, then $\sqrt[3]{3P(P^2+3P+3)+1}$
$\sqrt[3]{3999({\left(999\right)}^2+3\left(999\right)+3)+1}$
$\sqrt[3]{3{\left(999\right)}^3+3{\left(999\right)}^2+3\left(999\right)+1}$
$\sqrt[3]{3{(999+1)}^3}$
$\sqrt[3]{3{\left(1000\right)}^3}$
${\left({1000}^3\right)}^{\frac{1}{3}}$
$\Rightarrow 1000$