SSC CGL Tier II 21 Sep 2014 Paper

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Question : 26 of 100

Marks: +1, -0

If p = 99, then the value of

p (p^{2}+3 p+3 )

is

10000000
999000
999999
990000

Solution:

(3)

p = 99

(Given)

∴p (p^{2}+3 p+3 )=p^{3}+3 p^{2}+

=p^{3}+3 p^{2}+3 p+1-1

=(p+1)^{3}-1=(99+1)^{3}-1

=(100)^{3}-1=999999

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