SSC CGL 26 Oct 2014 Shift 1 Paper

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Question : 126 of 200

Marks: +1, -0

If p = 99, then the value of p (

p^2

+ 3p + 3) is

10000001
9999999
1000001
999999

Solution:

p = 99
∴ p (

p^2

+ 3p + 3)
=

p^3 + 3p^2

+ 3p
= (

p^3 + 3p^2

+ 3p + 1) – 1
=

(p + 1)^3

– 1 =

(99 + 1)^3

– 1
=

(100)^3

– 1 = 1000000 – 1
= 999999

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