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OG Quant Review - PS - Q 103

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by arvysri » Sat Jan 18, 2014 11:45 pm
Hello All,

I would like to listen to some comments and find alternate way(s) to approach the following question. If this question has been explained already in the forum, I apologize for the re-post.

OG Quant Review 2ED - PS - Question 103

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 ?

I tried to expand x^3-x trying to find a way to solve but ended up nowhere. Hence I switched to this route.

The question stem states the Left Hand Side(LHS) = Right Hand Side(RHS) and also mentions "for all numbers x".

So I set x=2 (some random number) and arrived here:

8-2 = (2-a) (2-b) (2-c)
6 = (2-a) (2-b) (2-c)

From here - three numbers when multiplied together gives 6 (3x2x1) and a>b>c.

I needed a 3, 2 and 1 and picked numbers for a,b and c.

6 = (2-1) (2-0) (2-(-1))

I believe my approach does not follow a set rule, might take extra time and break in a few scenarios(that I couldn't think of now). Please explain how to solve this question in a systematic manner.

Thank you!

Best Regards,
Arvind.
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by Uva@90 » Sun Jan 19, 2014 12:35 am
Hi Arvind,
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 ?
x^3 - x can be re-written as x(x^2-1) => x(x+1)(x-1)

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

since a>b>c , there fore value of (x-b) will be between (x-a) and (x-c)

in LHS of the above equation we can see that (x+1)>x>(x-1)

so x = x-b
hence [spoiler]b=0[/spoiler]

Hope it helps you.

Regards,
Uva.
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by GMATGuruNY » Sun Jan 19, 2014 4:12 am
arvysri 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 ?

-3
-1
0
1
3
Since a>b>c, (x-a)(x-b)(x-c) represents 3 factors in ASCENDING order, where (x-b) is the value of the MIDDLE factor.
Your approach -- to plug in a value for x and solve for b -- could lead to issues here.

Let x=3.
Then x³ - x = 3³ - 3 = 24.
Here, the middle factor = 3-b.

Two options for the 3 factors in ascending order:
Case 1: 1*4*6
Here, 3-b=4, implying that b=-1 (answer choice B).
Case 2: 2*3*4
Here, 10-b=3, implying that b=0 (answer choice C).
A test-taker who considers only Case 1 would be led to select answer choice B, which is incorrect.

For this reason, algebra is safer here:
x³ - x = x(x²-1) = x(x+1)(x-1).
Rearranging the factors in ascending order, we get:
(x-1)(x)(x+1).
Since the middle factor is x:
x-b = x
b=0.

The correct answer is C.
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by ygcrowanhand » Mon Apr 24, 2017 7:03 am
Hi GMATters,

Here's my video explanation of the problem: https://youtu.be/2Y6Yo5kUBhs

Enjoy!

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by Jeff@TargetTestPrep » Mon Dec 18, 2017 5:21 pm
arvysri 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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OG Quant Review - PS - Q 103

by ceilidh.erickson » Tue Dec 19, 2017 1:18 pm
arvysri wrote:Hello All,

I would like to listen to some comments and find alternate way(s) to approach the following question. If this question has been explained already in the forum, I apologize for the re-post.

OG Quant Review 2ED - PS - Question 103

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 ?

I tried to expand x^3-x trying to find a way to solve but ended up nowhere. Hence I switched to this route.

The question stem states the Left Hand Side(LHS) = Right Hand Side(RHS) and also mentions "for all numbers x".

So I set x=2 (some random number) and arrived here:

8-2 = (2-a) (2-b) (2-c)
6 = (2-a) (2-b) (2-c)

From here - three numbers when multiplied together gives 6 (3x2x1) and a>b>c.

I needed a 3, 2 and 1 and picked numbers for a,b and c.

6 = (2-1) (2-0) (2-(-1))

I believe my approach does not follow a set rule, might take extra time and break in a few scenarios(that I couldn't think of now). Please explain how to solve this question in a systematic manner.

Thank you!

Best Regards,
Arvind.
It helps to recognize that x^3 - x is GMAT code for the product of 3 consecutive integers. As Jeff pointed out, we can factor it as (x - 1)(x)(x + 1). This structure (or variations on it) have shown up in a number of official questions testing the properties of consecutive products.

For example, look at problem DS #170 in OG12:
If n is a positive integer, is n^3 – n divisible by 4?

(1) n = 2k + 1, where k is an integer
(2) n2 + n is divisible by 6
Here are a few other similar examples:
https://www.beatthegmat.com/x-x-1-x-k-t ... tml#770020
https://www.beatthegmat.com/totaly-lost ... tml#716315
https://www.beatthegmat.com/is-x-x-2-x- ... tml#718646
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Harvard Graduate School of Education
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by Brent@GMATPrepNow » Tue Jul 24, 2018 8:28 am
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
Given: x³ - x = (x - a)(x - b)(x - c)
Factor x from left side: x(x² - 1) = (x - a)(x - b)(x - c)
x² - 1 is a difference of squares, so we can factor that: x(x + 1)(x - 1) = (x - a)(x - b)(x - c)
Rewrite first 2 terms as: (x - 0)(x - -1)(x - 1) = (x - a)(x - b)(x - c)
Rearrange as: (x - 1)(x - 0)(x - -1) = (x - a)(x - b)(x - c)

Since, we're told that c < b < a, we can see that a = 1, b = 0 and c = -1

Answer: C

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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