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Re: If a, b, and c are constants, a > b > c, and x^3 - x = (x - a)(x - b)
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02 Mar 2015, 19:18

8

1

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....

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

Re: If a, b, and c are constants, a > b > c, and x^3 - x = (x - a)(x - b)
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07 Feb 2016, 19:39

1

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 .

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

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?

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

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

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

Re: If a, b, and c are constants, a > b > c, and x^3 - x = (x - a)(x - b)
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20 Jul 2017, 16:36

2

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
_________________

Jeffery Miller Head of GMAT Instruction

GMAT Quant Self-Study Course 500+ lessons 3000+ practice problems 800+ HD solutions

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

Answer: C

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? & what about the negative?

If a, b, and c are constants, a > b > c, and x^3 - x = (x - a)(x - b)
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31 Mar 2018, 08:08

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....