If p = 999, then the value of \sqrt^3{p(p^2 + 3p + 3) +1} is

RRB NTPC Mains or Stage 2 Model Paper 8 with solutions

Question : 17 of 120

Marks: +1, -0

If p = 999, then the value of

√^3{p(p^2 + 3p + 3) +1}

Solution:

(a)

p = 999

(Given)
Expression

⇒ √^3{ 3p^+3p^2+ 3p +1}

⇒ √^3( p +1)^3

⇒ p + 1 = 999 + 1 = 1000

Go to Question:

Prev Question

Next Question