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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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EMPOWERgmat Instructor
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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, 19:18
8
KUDOS
Expert's post
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

[Reveal] Spoiler:
C

GMAT assassins aren't born, they're made,
Rich
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# Rich Cohen

Co-Founder & GMAT Assassin

Special Offer: Save $75 + GMAT Club Tests Free Official GMAT Exam Packs + 70 Pt. Improvement Guarantee www.empowergmat.com/ ***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*********************** Board of Directors Joined: 17 Jul 2014 Posts: 2753 Location: United States (IL) Concentration: Finance, Economics GMAT 1: 650 Q49 V30 GPA: 3.92 WE: General Management (Transportation) Re: If a, b, and c are constants, a > b > c, and x^3 - x = (x - a)(x - b) [#permalink] ### Show Tags 16 Nov 2015, 20:16 i solved it this way: x^3 - x = x(x+1)(x-1)=(x-a)(x-b)(x-c) I did not try to solve which one is bigger/smaller x = x-b always, since C is the middle number for x=x-b to be true, b must be 0. Manager Joined: 01 Aug 2014 Posts: 58 Schools: Rotman '17 (A) GMAT 1: 710 Q44 V42 Re: If a, b, and c are constants, a > b > c, and x^3 - x = (x - a)(x - b) [#permalink] ### Show Tags 07 Feb 2016, 19:31 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 Current Student Joined: 20 Mar 2014 Posts: 2685 Concentration: Finance, Strategy Schools: Kellogg '18 (M) GMAT 1: 750 Q49 V44 GPA: 3.7 WE: Engineering (Aerospace and Defense) Re: If a, b, and c are constants, a > b > c, and x^3 - x = (x - a)(x - b) [#permalink] ### Show Tags 07 Feb 2016, 19:39 1 This post received KUDOS 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. Intern Joined: 01 Jul 2016 Posts: 3 Re: If a, b, and c are constants, a > b > c, and x^3 - x = (x - a)(x - b) [#permalink] ### Show Tags 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! EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 11240 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: 340 Q170 V170 Re: If a, b, and c are constants, a > b > c, and x^3 - x = (x - a)(x - b) [#permalink] ### Show Tags 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). GMAT assassins aren't born, they're made, Rich _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com # Rich Cohen Co-Founder & GMAT Assassin Special Offer: Save$75 + GMAT Club Tests Free
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Manager
Joined: 09 Aug 2016
Posts: 68
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 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
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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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24 Apr 2017, 07:47
Hi GMATters,

Here's my video explanation of the problem:

Enjoy!

Rowan
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PS have you seen the new GMAT Work and Rates guide? Comes with a free 8-video course.

https://yourgmatcoach.podia.com/courses/how-to-beat-gmat-work-and-rates-problems

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Joined: 15 Dec 2016
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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 Jul 2017, 15:33
metallicafan wrote:
ezhilkumarank wrote:
Quote:
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?

We are not assuming anything. We are just equating the left hand side to the right hand side of the equation.

Any other thoughts?????

Yes, as I said, for example, 6*2=3*4, but we cannot say that 6 = 3. Based on this, we cannot say that (x-0)=(x-b)
Well, I am studying too much, probably I am seeing things that probably don't really matter LOL.

Hi in your example --- 6 * 2 * 1 = 3 *4 * 1 will not fit in the equation (x-0) (x+1) (x-1) = (x +a) (x+b) (x+C) with a>b>c ... put in values and see if it works ...here are some of my attempts ..

hence not an applicable test

trying random values, assuming X = 5 and
a = 3
b = 2
c = 1

that is 5 (x-0) * 6 (x+1) * 4 (x-1) = 8 (x+a) * 7 (x+b) * 6 (x+c)

not working

trying random values, assuming X = 5 and
a = -1
b = -2
c = -3

that is 5 (x-0) * 6 (x+1) * 4 (x-1) = 4 (x+a) * 3 (x+b) * 2 (x+c)

not feasible ...

only way this is possible is the following

assuming X = 5 and
a = 1
b = 0
c = -1

that is 5 (x-0) * 6 (x+1) * 4 (x-1) = 6 (x+a) * 5 (x+b) *4 (x+c)....working

Whatever the value of x but with a = 1 | b = 0 | c = -1 ...below equation will hold ..

(x-0) (x+1) (x-1) = (x +a) (x+b) (x+C)

hence 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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20 Jul 2017, 16:36
1
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Expert's post
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.

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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 Oct 2017, 12:00
JeffTargetTestPrep wrote:
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.

Silly question, but how do we get algebraically get to the 0? Is it because we are subtracting x from both sides? Why is it not replaced with 1?

(x - b) = x

b = 0

Thanks!
If a, b, and c are constants, a > b > c, and x^3 - x = (x - a)(x - b)   [#permalink] 19 Oct 2017, 12:00

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