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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]  19 Feb 2014, 00:19
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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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Kudos [?]: 42299 [1] , given: 6012

Re: If a, b, and c are constants, a > b > c, and x^3 - x = (x - [#permalink]  19 Feb 2014, 00:19
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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 $$c=-1$$.

Answer: C.
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Re: If a, b, and c are constants, a > b > c, and x^3 - x = (x - [#permalink]  19 Feb 2014, 01:54
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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
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Re: If a, b, and c are constants, a > b > c, and x^3 - x = (x - [#permalink]  22 Feb 2014, 05:03
Expert's post
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]  01 May 2014, 07:27
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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.

Answer is C
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Re: If a, b, and c are constants, a > b > c, and x^3 - x = (x - [#permalink]  17 Jul 2014, 00:07
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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

Answer = C
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Re: If a, b, and c are constants, a > b > c, and x^3 - x = (x - [#permalink]  17 Jul 2014, 04:17
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]  17 Jul 2014, 04:43
Expert's post
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]  17 Jul 2014, 17:52
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]  02 Mar 2015, 12:11
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]  02 Mar 2015, 18:18
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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

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] 02 Mar 2015, 18:18
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# If a, b, and c are constants, a > b > c, and x^3 - x = (x -

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