If the
roots of the equation x

^{3}-ax^{2}+bx-c=0 are three consecutive integers, then what is the smallest possible value of b?
(A)-1/√3
(B)-1 (C)0
(4)1 (5)1/√3

Solution follows here:

Solution:

This problem is a typical simplification type.

x

^{3}-ax^{2}+bx-c=0 ---(1)
Let the three routes be n-1, n, n+1: Hence these values
satisfy the equation (1).

(n-1)

^{3}-a(n-1)^{2}+b(n-1)-c=0 ---(2)
n

^{3}-an^{2}+bn-c=0 ---(3)
(n+1)

^{3}-a(n+1)^{2}+b(n+1)-c=0 ---(4)
(3) – (2) yields => n

^{3}-(n-1)^{3}-a(n^{2}-(n-1)^{2})+b(n-(n-1)) = 0
=> 1-3n+3n

^{2}-a(2n-1)+b = 0 ---(5)
(4) – (3) yields => (n+1)

^{3}-n^{3}-a((n+1)^{2}-n^{2})+b(n+1-n) = 0
=> 1+3n+3n

^{2}-a(2n+1)+b = 0 ---(6)
Multiply (5) with (2n+1) and (6) with (2n-1):

(2n+1)(1-3n+3n

^{2}) - a(2n+1) (2n-1) + b(2n+1) = 0 ---(7)
(2n-1)( 1+3n+3n

^{2}) - a(2n+1) (2n-1) + b(2n-1) = 0 ---(8)
(8) – (7) yields:

6n

^{2}-2+b(-2) = 0 => -2b = 2-6n^{2}^{ }=>**b = 3n**^{2}-1
n

^{2}is always positive, and its minimum value is zero.
Hence minimum value of b = 3(0)-1 =

**-1****Answer (B)**

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