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If a, b, and c are constants, a > b > c, and x^3 - x = (x -
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02 Jun 2020, 02:02
02 Jun 2020, 02:02
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If a, b, and c are constants, \(a > b > c\), and \(x^3 - x = (x - a)(x - b)(x - c)\) for all numbers x, what is the value of b ?
(A) - 3
(B) -1
(C) 0
(D) 1
(E) 3
(A) - 3
(B) -1
(C) 0
(D) 1
(E) 3
Re: If a, b, and c are constants, a > b > c, and x^3 - x = (x -
[#permalink]
02 Jun 2020, 02:02
02 Jun 2020, 02:02
Expert Reply
Post A Detailed Correct Solution For The Above Questions And Get A Kudos.
Question From Our New Project: GRE Quant Challenge Questions Daily - NEW EDITION!
Question From Our New Project: GRE Quant Challenge Questions Daily - NEW EDITION!
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Re: If a, b, and c are constants, a > b > c, and x^3 - x = (x -
[#permalink]
02 Jun 2020, 04:57
02 Jun 2020, 04:57
3
Carcass wrote:
If a, b, and c are constants, \(a > b > c\), and \(x^3 - x = (x - a)(x - b)(x - c)\) for all numbers x, what is the value of b ?
(A) - 3
(B) -1
(C) 0
(D) 1
(E) 3
(A) - 3
(B) -1
(C) 0
(D) 1
(E) 3
Given: x³ - x = (x - a)(x - b)(x - c)
Factor x from left side: x(x² - 1) = (x - a)(x - b)(x - c)
x² - 1 is a difference of squares, so we can factor that: x(x + 1)(x - 1) = (x - a)(x - b)(x - c)
Rewrite first 2 terms as: (x - 0)(x - -1)(x - 1) = (x - a)(x - b)(x - c)
Rearrange as: (x - 1)(x - 0)(x - -1) = (x - a)(x - b)(x - c)
Since, we're told that c < b < a, we can see that a = 1, b = 0 and c = -1
Answer: C
Cheers,
Brent
