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Each week we'll be posting several questions from The Official Guide For GMAT® Quantitative Review, 2ND Edition and then after couple of days we'll provide Official Answer (OA) to them along with a slution.

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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. Substituting these values you get the values of a, b, and c. _________________

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

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02 Mar 2015, 13:11

Bunuel wrote:

bankerboy30 wrote:

How do we know so is equal the numbers are -1,0,1. I know there consecutive but why not 2,3,4?

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. Substituting these values you get the values of a, b, and c.

I still dont understand this logic. Are we simply looking for any value of X that makes the entire equation 0? What do you specifically mean by "as x^3.... is true for all x-es" - I really dont understand this.

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 - [#permalink]

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03 May 2016, 08:50

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If a, b, and c are constants, a > b > c, and x^3 - x = (x - [#permalink]

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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?

Thanks in advance!

gmatclubot

If a, b, and c are constants, a > b > c, and x^3 - x = (x -
[#permalink]
21 Jul 2016, 17:53

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