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

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

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I follow the official explanation through:

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

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.

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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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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


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 - a)(x - b) [#permalink]

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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

OA is
[Reveal] Spoiler:
C


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

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!

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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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metallicafan wrote:
Bunuel, could you explain this with more detail?

Bunuel wrote:
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 \(c\).
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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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New post 12 Aug 2013, 19:48
Hi,

I realize I'm a bit late to the discussion, but have this question.

Why couldn't you do the following:

Take the equation x^3-x=(x-a)(x-b)(x-c) and rearrange it like so:

(x^3-x)/((x-a)(x-c))=x-b

and then, since this equation should hold true for all numbers x, plug in x=1, which would give you:

0/((1-a)(1-c))=1-b

The left side of the equation goes to 0, giving

0=1-b, thus

b=1

Am I misunderstanding what is meant by "for all numbers x"? Thanks for your help!

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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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New post 13 Aug 2013, 02:01
zx8009 wrote:
Hi,

I realize I'm a bit late to the discussion, but have this question.

Why couldn't you do the following:

Take the equation x^3-x=(x-a)(x-b)(x-c) and rearrange it like so:

(x^3-x)/((x-a)(x-c))=x-b

and then, since this equation should hold true for all numbers x, plug in x=1, which would give you:

0/((1-a)(1-c))=1-b

The left side of the equation goes to 0, giving

0=1-b, thus

b=1

Am I misunderstanding what is meant by "for all numbers x"? Thanks for your help!


You cannot divide by (x-a)(x-c), since (x-a)(x-c) can be zero and division by zero is not allowed.

Hope it's clear.
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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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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
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If a, b, and c are constants, a > b > c, and x^3 - x = (x - a)(x - b) [#permalink]

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

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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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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 - a)(x - b) [#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 - a)(x - b) [#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 - a)(x - b) [#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 - a)(x - b) [#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 - a)(x - b) [#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 - a)(x - b) [#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 - a)(x - b) [#permalink]

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New post 23 Oct 2014, 21:58
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)

X (X^2-1) = (X-A) (X-B) (X-C)

X (X-1) (X+1) = (X-A) (X-B) (X-C)

Since, A > B> C, Therefore, (X-A) < (X-B) < (X-C)

(X-1) X (X+1) = (X-A) (X-B) (X-C)

(X-A) = (X-1) ------------ A = 1
(X-B) = X ----------------- B = 0
(X-C) = (X+1) ----------- C = -1

Therefore Answer is OA C

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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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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 - a)(x - b)   [#permalink] 02 Mar 2015, 13:11

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