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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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26 Jul 2010, 11:01
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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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26 Jul 2010, 11:05
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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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26 Jul 2010, 11:14
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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 $$c=-1$$.

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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, 13:44
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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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15 Aug 2010, 17:28
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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!

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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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16 Aug 2010, 03:23
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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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12 Aug 2013, 18:48
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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13 Aug 2013, 01:01
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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13 Aug 2013, 02:59
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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
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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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19 Feb 2014, 00:19
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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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19 Feb 2014, 00:19
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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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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 - a)(x - b) [#permalink]

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22 Feb 2014, 05:03
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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 - a)(x - b) [#permalink]

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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.

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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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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

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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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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 - a)(x - b) [#permalink]

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17 Jul 2014, 04:43
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 - a)(x - b) [#permalink]

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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 - a)(x - b) [#permalink]

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23 Oct 2014, 20:58
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

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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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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 - a)(x - b)   [#permalink] 02 Mar 2015, 12:11

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