SSC CGL 7 Sep 2016 Paper

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Question : 62 of 100

Marks: +1, -0

If p = 99, then the value of p (

p^2

+ 3p + 3) will be

999999
1000000
1000001
999998

Solution:

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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