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Question:

If the roots of the equation x³− ax² + bx – c = 0 are three consecutive integers, then what is the smallest possible value of b?

Explanation:

If p, q and r are the roots of a cubic equation ax³ + bx² + cx + d = 0,

then pq + pr + qr = c/a

If a is 1, then pq + qr + pr = c

Comparing the equation ax³ + bx² + cx + d = 0 with the equation in the question x³− ax² + bx – c = 0, we get

pq + qr + pr = b

Let the three roots of the given equation be (n – 1), n and (n + 1).

∴ (n – 1)n + n(n + 1) + (n – 1)(n + 1) = b

∴ n² – n + n² + n + n² – 1 = b

∴ 3n² – 1 = b

∵ n² ≥ 0, minimum value of b occurs at n = 0

∴ Minimum value of b = –1

Hence, option (b).