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

Question

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

kelvinp · Accepted Answer

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

Bunuel · Answer

Algebra Theory

Theory   
 Math : Algebra 101 
 Algebra: Tips and hints 
 FOIL on the GMAT: Simplifying and Expanding 
 Factoring Quadratics 
 Solving Quadratic Equations  
 Using Algebra vs. Logic on GMAT Quant Questions

Bunuel · Answer

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.

metallicafan · Answer

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 
 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?

jpr200012 · Answer

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!

Bunuel · Answer

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 .

zx8009 · Answer

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!

Bunuel · Answer

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.

Bunuel · Answer

Algebra Questions

Tough Algebra Set 
 Algebra Ability Quiz by e-GMAT 
 DS Questions 
 PS Questions

AKG1593 · Answer

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)
 
Therefore,a=1
b=0
c=-1

PareshGmat · Answer

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

EMPOWERgmatRichC · Answer

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:  C

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

Anonamy · Answer

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

ENGRTOMBA2018 · Answer

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

Schnauss · Answer

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!

EMPOWERgmatRichC · Answer

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

Ndkms · Answer

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

PGTLrowanhand · Answer

Hi GMATters,

Here's my video explanation of the problem:

2Y6Yo5kUBhs

Enjoy!

Rowan

JeffTargetTestPrep · Answer

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

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

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

GMAT Problem Solving (PS) Questions