If P = 99, then the value of P($$P^{2}$$ + 3P + 3) is
P($$P^{2}$$ + 3P + 3)Â
= P($$P^{2} + 4P + 3-P$$)
= P[$$(P+1)(P+3)-P)$$]
Given that P=99, Substituting it in above equation, we get
= 99[$$(99+1)(99+3)-99$$]
=99($$10200-99$$)
=99(10101)=999999
Create a FREE account and get:
Detailed syllabus & Topic-wise Weightage
By proceeding you agree to create your account
Free CAT Syllabus PDF will be sent to your email address soon !!!
