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Difficulty: 505-555 Level,    Algebra,                      
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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


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]
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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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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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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\(x^3 - x = (x - a)(x - b)(x - c)\)

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

(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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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:

GMAT assassins aren't born, they're made,
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Re: If a, b, and c are constants, a > b > c, and x^3 - x = (x - a)(x - b) [#permalink]
There is some bit of knowledge I'm missing. How are we supposed to innately know that (x-a)(x-b)(x-c)=(x+1)(x-0)(x-1)?

I'm even more confused after reading through these approaches. When I attempted to 'pull' x out of x^3-x I got x^2(x−1) which is different that the approach I'm seeing here.

Thanks in advance
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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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Anonamy wrote:
There is some bit of knowledge I'm missing. How are we supposed to innately know that (x-a)(x-b)(x-c)=(x+1)(x-0)(x-1)?

I'm even more confused after reading through these approaches. When I attempted to 'pull' x out of x^3-x I got x^2(x−1) which is different that the approach I'm seeing here.

Thanks in advance


You need to practice algebra a bit more.

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

So once you are able to get to \(x^3-x = x*(x+1)(x-1)\), you are now set to compare it with \((x-a)(x-b)(x-c)\)

Thus, you get \(x*(x+1)(x-1) = (x-a)(x-b)(x-c)\) ---> \((x-0)(x+1)(x-1) = (x-a)(x-b)(x-c)\) ---> \((x-0)(x- (-1))(x-1) = (x-a)(x-b)(x-c)\), giving you 3 values of -1,0,1 .

As a<b<c, b can only be = 0.

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

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 - a)(x - b) [#permalink]
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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,
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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 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
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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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Hi GMATters,

Here's my video explanation of the problem:



Enjoy!

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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


Let’s simplify the given equation:

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)

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

Since a > b > c:

(x - b) = x

b = 0

Alternate Solution:

If we let x = a in the equation, we get a^3 - a = 0; i.e., a^3 = a. Similarly, letting x = b and x = c, we get b^3 = b and c^3 = c. We know that the only three numbers whose cube is equal to the number itself are 1, 0, and -1. We have a > b > c; therefore, we must have a = 1, b = 0, and c = -1.

Answer: C
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