TS Police Constable Telugu Model Paper 4

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Question : 55 of 200

Marks: +1, -0

p=99,అయిన

p(p^2+3p+3)

విలువ ?

999999
1000000
10000001
999998

Solution:

p(p^2+3p+3)=p^3+3p^2+3p

=p^3+3p^2+3p+1-1=(p+1)^3-1

⇒(p+1)^3-1

⇒(99+1)^3-1=1000000-1=999999

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