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# If a, b, and c are constants, a > b > c, and x^3 - x = (x -

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

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19 Feb 2014, 01:19
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The Official Guide For GMAT® Quantitative Review, 2ND Edition

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

Problem Solving
Question: 103
Category: Algebra Simplifying algebraic expressions
Page: 75
Difficulty: 650

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

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19 Feb 2014, 01:19
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SOLUTION

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

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:

$$x=0$$ --> $$x^3-x=(x-a)(x-b)(x-c)$$ becomes: $$0=-abc$$, so one of the unknowns equals to zero;

$$x=1$$ --> $$x^3-x=(x-a)(x-b)(x-c)$$ becomes: $$0=(1-a)(1-b)(1-c)$$, so one of the unknowns equals to 1;

$$x=-1$$ --> $$x^3-x=(x-a)(x-b)(x-c)$$ becomes: $$0=(-1-a)(-1-b)(-1-c)$$, so one of the unknowns equals to -1.

As $$a > b > c$$ then $$a=1$$, $$b=0$$ and $$c=-1$$.

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

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19 Feb 2014, 02:54
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Option C:
x^3-x can be re-written as:(x-1)*x*(x+1) in ascending order.
(x-1)*x*(x+1)=(x-a)*(x-b)*(x-c)
[As a>b>c=>-a<-b<-c => x-a<x-b<x-c]
Therefore,a=1
b=0
c=-1
Math Expert
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Posts: 39589
Re: If a, b, and c are constants, a > b > c, and x^3 - x = (x - [#permalink]

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22 Feb 2014, 06:03
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SOLUTION

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

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:

$$x=0$$ --> $$x^3-x=(x-a)(x-b)(x-c)$$ becomes: $$0=-abc$$, so one of the unknowns equals to zero;

$$x=1$$ --> $$x^3-x=(x-a)(x-b)(x-c)$$ becomes: $$0=(1-a)(1-b)(1-c)$$, so one of the unknowns equals to 1;

$$x=-1$$ --> $$x^3-x=(x-a)(x-b)(x-c)$$ becomes: $$0=(-1-a)(-1-b)(-1-c)$$, so one of the unknowns equals to -1.

As $$a > b > c$$ then $$a=1$$, $$b=0$$ and $$c=-1$$.

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

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01 May 2014, 08:27
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Alternative SOLUTION
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

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

Using $$(a-b)^2 = (a-b)(a+b)$$ algebra form, the expression becomes $$( x ) (x - 1)(x + 1) = (x-a)(x-b)(x-c)$$

Effectively, the solutions are 0, -1, and +1.
As we are given that a > b > c, the order is therefore +1 > 0 > -1
B is therefore 0.

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

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17 Jul 2014, 01:07
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$$x^3 - x = (x - a)(x - b)(x - c)$$

$$x (x^2 - 1) = (x - a)(x - b)(x - c)$$

(x + 0) (x+1) (x-1) = (x - a)(x - b)(x - c)

Given that a>b>c; so reorganising LHS of the equation

(x-1) (x + 0) (x+1) = (x - a)(x - b)(x - c)

b = 0

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

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17 Jul 2014, 05:17
How do we know so is equal the numbers are -1,0,1. I know there consecutive but why not 2,3,4?
Math Expert
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Posts: 39589
Re: If a, b, and c are constants, a > b > c, and x^3 - x = (x - [#permalink]

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17 Jul 2014, 05:43
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.
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Re: If a, b, and c are constants, a > b > c, and x^3 - x = (x - [#permalink]

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17 Jul 2014, 18:52
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?

Expanding $$x^3 - x$$ gives the values -1, 0, 1 besides x. Those are derived as shown in my above post

$$(x+2)(x+3)(x+4) \neq{x^3 - x}$$
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Re: If a, b, and c are constants, a > b > c, and x^3 - x = (x - [#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.
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Re: If a, b, and c are constants, a > b > c, and x^3 - x = (x - [#permalink]

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02 Mar 2015, 19:18
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Hi erikvm,

This is a "layered" concept and it's easy to get "lost" in this prompt because you're used to solving for the final values in most Quant questions.

Here, the 3 "final" numbers are (X - A), (X - B) and (X - C), but the question is NOT asking for any of the 3 final numbers...it's asking for a "piece" of one of them....the value of B.

To answer it, you have to ignore the A, B and C for a moment and go back to the prior "term"

X^3 - X

This can be factored down into 3 pieces. Here's how...

X^3 - X

First, factor out an X...

(X)(X^2 - 1)

Next, reverse-FOIL the other term....
(X)(X+1)(X-1)

Since we're multiplying 3 terms, it doesn't matter what the order is. I'm going to put them in order from least to greatest...

(X-1)(X)(X+1)

Now, looking at THIS, you can figure out what A, B and C are. Since A>B>C, then....

A = +1
B = 0
C = -1

[Reveal] Spoiler:
C

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# Special Offer: Save $75 + GMAT Club Tests 60-point improvement guarantee www.empowergmat.com/ ***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*********************** GMAT Club Legend Joined: 09 Sep 2013 Posts: 15918 Re: If a, b, and c are constants, a > b > c, and x^3 - x = (x - [#permalink] ### Show Tags 03 May 2016, 08:50 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ Intern Joined: 01 Jul 2016 Posts: 3 If a, b, and c are constants, a > b > c, and x^3 - x = (x - [#permalink] ### Show Tags 21 Jul 2016, 17:53 EMPOWERgmatRichC wrote: Hi erikvm, This is a "layered" concept and it's easy to get "lost" in this prompt because you're used to solving for the final values in most Quant questions. Here, the 3 "final" numbers are (X - A), (X - B) and (X - C), but the question is NOT asking for any of the 3 final numbers...it's asking for a "piece" of one of them....the value of B. To answer it, you have to ignore the A, B and C for a moment and go back to the prior "term" X^3 - X This can be factored down into 3 pieces. Here's how... X^3 - X First, factor out an X... (X)(X^2 - 1) Next, reverse-FOIL the other term.... (X)(X+1)(X-1) Since we're multiplying 3 terms, it doesn't matter what the order is. I'm going to put them in order from least to greatest... (X-1)(X)(X+1) Now, looking at THIS, you can figure out what A, B and C are. Since A>B>C, then.... A = +1 B = 0 C = -1 Final Answer: [Reveal] Spoiler: C GMAT assassins aren't born, they're made, Rich What I found confusing about this problem is that we are not told that the left side of the equation is equal to 0 (x^3-x=0), so how do we know to begin factoring x^3-x here into x(x-1)(x=1)? Am I missing something really basic here? Thanks in advance! EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 9252 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: 340 Q170 V170 Re: If a, b, and c are constants, a > b > c, and x^3 - x = (x - [#permalink] ### Show Tags 24 Dec 2016, 10:16 Hi Schnauss, The 'act' of factoring does not require an equation - it's just a way of re-writing information that you've been given. For example, if you're told that you have 'two boxes that each contain the same number of widgets', then you can write that as (2)(W)... even though you don't have an actual equation yet. If I wanted to, I could rewrite 2W as (W + W). Certain GMAT questions just come down to organizing information in a way that makes it easy to answer the given question, so you should think about how you choose to take notes and 'translate' sentences ('your way' might not be the only way, and there might be additional 'steps' that you can take to simplify what you have written). GMAT assassins aren't born, they're made, Rich _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com # Rich Cohen Co-Founder & GMAT Assassin # Special Offer: Save$75 + GMAT Club Tests

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

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26 Jan 2017, 07:51
If you dont know the algebra for this most probably you cannot solve this question.

x^3 - x = x(x^2 - 1) . GMAT algebra fundamentals require you to know/memorise that x^2 - 1 = (x-1) * (x+1)

So long story short: x^3 - x = x * (x-1) * (x+1) = (x - a)(x - b)(x - c)

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

We know that a > b > c

Hence (x-c) > (x-b) > (x-a) . This is because in this question x is a constant rather than a variable. So the maximum value of any of the three operation, (x-c), (x-b) or (x-a) is the one that uses the SMALLEST value (i.e. c):

Then you just do the mapping

out of x, (x-1) and (x+1) the (x+1) has the GREATEST value. Hence (x+1) = (x - c) => x+1 = x -c which gives c = -1
out of x, (x-1) and (x+1) the (x-1) has the SMALLEST value . Hence (x-1) = (x - a) which gives a = 1

therefore the only case that left is the x = (x-b) which gives b = 0
Re: If a, b, and c are constants, a > b > c, and x^3 - x = (x -   [#permalink] 26 Jan 2017, 07:51
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