Question 66

If P = 99, then the value of P($$P^{2}$$ + 3P + 3) is

Solution

P($$P^{2}$$ + 3P + 3) 

= 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


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