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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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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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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 \(c=-1\).

Answer: 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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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
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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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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 \(c=-1\).

Answer: 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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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.

Answer is 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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\(x^3 - x = (x - a)(x - b)(x - c)\)

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

(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

Answer = 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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New post 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?
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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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New post 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?


Please check here: if-a-b-and-c-are-constants-a-b-c-and-x-3-x-x-167671.html#p1335360

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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New post 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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New post 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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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


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

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New post 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!
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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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New post 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).

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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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New post 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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