If P = 999, then cube root of√P(P^2+3P+3)+1=(a) 1000 (b) 999 (c) 1002 (d) 998

Question

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

(a) 1000 

(b) 999 

(c) 1002 

(d) 998

Answer 1

Correct option is (a) 1000

P = 999, then 

\(\sqrt[3]{P(P^2 + 3P + 3) + 1}\)

\(= \sqrt[3]{999((999)^2 + 3(999) + 3) + 1}\)

\(= \sqrt[3]{(999)^3 + 3(999)^2 + 3+(999)+ 1}\)

\(= \sqrt[3]{(999+ 1)^3}\)

\(= \sqrt[3]{(1000)^3}\)

\(= {(1000^3)^\frac 13}\)

\(=1000\)

Answer 2

The answer is: (a) 1000