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Each week we'll be posting several questions from The Official Guide For GMAT® Quantitative Review, 2ND Edition and then after couple of days we'll provide Official Answer (OA) to them along with a slution.

We'll be glad if you participate in development of this project: 1. Please provide your solutions to the questions; 2. Please vote for the best solutions by pressing Kudos button; 3. Please vote for the questions themselves by pressing Kudos button; 4. Please share your views on difficulty level of the questions, so that we have most precise evaluation.

Then the explanation substitutes in \((x-0)(x-1)(x+1)\) on the right side. This last step seems a little confusing, but substituting in numbers verifies it works.

DON'T READ THIS UNTIL YOU SOLVE IT. The solution in the OG is the following: \((x-a)(x-b)(x-c) = x^3-x\) But \(x^3-x\) is (x-0)(x-1)(x+1), then a, b, and c are 0, 1 and -1 in some order. Since a>b>c, it follows that a=1, b=0, and c=-1.

However, why could we assume that (x-0)=(x-b), (x+1)=(x-c) and (x-1)=(x-a)? Probably, x(x-1)(x+1) are different factors which produce the same result as (x-a)(x-b)(x-c) do. For example, 6*2=3*4, but we cannot say that 6 = 3. Am I missing something? _________________

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Re: If a, b, and c are constants, a > b > c, and x^3 - x = (x - a)(x - b) [#permalink]

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15 Aug 2010, 18:28

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\(x^3 - x = (x-a)(x-b)(x-c)\) \(x(x^2-1)=(x-a)(x-b)(x-c)\) \(x(x+1)(x-1)=(x-a)(x-b)(x-c)\)

We know that \(a>b>c\). We also know that \(x+1>x>x-1\).

Now, rearrange the terms from highest to lowest on both sides.

\((x+1)x(x-1)=(x-c)(x-b)(x-a)\)

So you can see \(x+1 = x-c\), \(x = x-b\), \(x-1=x-a\).

The easiest one is \(x = x-b\). Obviously, \(b = 0\). The problem asks for b, so no further work required. However, you can get a or c easy if the problem had asked for them instead.

This is a tough problem. Makes you stop and think after not working it for a few days Just don't forget about the inequality. That was the key for me!

As \(x^3-x=(x-a)(x-b)(x-c)\) is true for ALL \(x-es\) than it must be true for \(x=0\), \(x=1\) and \(x=-1\) too:

Why must be true that \(x=0\), \(x=1\) and \(x=-1\)? \(x^3-x\) is not 0.

We are told that "\(x^3-x=(x-a)(x-b)(x-c)\) is true for ALL \(x-es\)" so this equation must also be true for \(x=0\), \(x=1\) and \(x=-1\). I chose these numbers as \(x^3-x\) (LHS) equals to zero for them, hence it's easy to find the possible values of \(a\), \(b\) and \(c\).
_________________

Re: If a, b, and c are constants, a > b > c, and x^3 - x = (x - a)(x - b) [#permalink]

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13 Aug 2013, 03:59

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jpr200012 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

x^3-x=(x-a)(x-b)(x-c) or, x (x^2 -1) = (x-a)(x-b)(x-c) or, x (x+1)(x-1) = (x-a)(x-b)(x-c) or, (x-1)(x-0){x-(-1)} = (x-a)(x-b)(x-c) Given, a > b > c Same, 1>0>-1 Finally, a=1 , b=0 and c = -1
_________________

As x^3-x=(x-a)(x-b)(x-c) is true for ALL x-es than it must be true for x=0, x=1 and x=-1 too. Substituting these values you get the values of a, b, and c.
_________________

Re: If a, b, and c are constants, a > b > c, and x^3 - x = (x - a)(x - b) [#permalink]

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02 Mar 2015, 13:11

Bunuel wrote:

bankerboy30 wrote:

How do we know so is equal the numbers are -1,0,1. I know there consecutive but why not 2,3,4?

As x^3-x=(x-a)(x-b)(x-c) is true for ALL x-es than it must be true for x=0, x=1 and x=-1 too. Substituting these values you get the values of a, b, and c.

I still dont understand this logic. Are we simply looking for any value of X that makes the entire equation 0? What do you specifically mean by "as x^3.... is true for all x-es" - I really dont understand this.