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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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26 Jul 2010, 11:01
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The Official Guide For GMAT® Quantitative Review, 2ND EditionIf 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 Problem Solving Question: 103 Category: Algebra Simplifying algebraic expressions Page: 75 Difficulty: 650 GMAT Club is introducing a new project: The Official Guide For GMAT® Quantitative Review, 2ND Edition  Quantitative Questions ProjectEach 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. We'll be glad if you participate in development of this project: 1. Please provide your solutions to the questions; 2. Please vote for the best solutions by pressing Kudos button; 3. Please vote for the questions themselves by pressing Kudos button; 4. Please share your views on difficulty level of the questions, so that we have most precise evaluation. Thank you!
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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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26 Jul 2010, 11:05
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I follow the official explanation through:
\(x^3x=(xa)(xb)(xc)\) \(x(x^21)=(xa)(xb)(xc)\) \(x(x+1)(x1)=(xa)(xb)(xc)\)
Then the explanation substitutes in \((x0)(x1)(x+1)\) on the right side. This last step seems a little confusing, but substituting in numbers verifies it works.



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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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26 Jul 2010, 11:14
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jpr200012 wrote: If a, b, and c are constants, \(a > b > c\), and \(x^3x=(xa)(xb)(xc)\) for all numbers \(x\), what is the value of \(b\)?
(A) 3 (B) 1 (C) 0 (D) 1 (E) 3 As \(x^3x=(xa)(xb)(xc)\) is true for ALL \(xes\) than it must be true for \(x=0\), \(x=1\) and \(x=1\) too: \(x=0\) > \(x^3x=(xa)(xb)(xc)\) becomes: \(0=abc\), so one of the unknowns equals to zero; \(x=1\) > \(x^3x=(xa)(xb)(xc)\) becomes: \(0=(1a)(1b)(1c)\), so one of the unknowns equals to 1; \(x=1\) > \(x^3x=(xa)(xb)(xc)\) becomes: \(0=(1a)(1b)(1c)\), so one of the unknowns equals to 1. As \(a > b > c\) then \(a=1\), \(b=0\) and \(c=1\). Answer: C.
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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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15 Aug 2010, 13:44
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If a, b, and c are constants, a>b>c, and \(x^3x=(xa)(xb)(xc)\) for all numbers x, what is the value of b? a. 3 b. 1 c. 0 d. 1 e. 3 OA is DON'T READ THIS UNTIL YOU SOLVE IT. The solution in the OG is the following: \((xa)(xb)(xc) = x^3x\) But \(x^3x\) is (x0)(x1)(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 (x0)=(xb), (x+1)=(xc) and (x1)=(xa)? Probably, x(x1)(x+1) are different factors which produce the same result as (xa)(xb)(xc) do. For example, 6*2=3*4, but we cannot say that 6 = 3. Am I missing something?
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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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15 Aug 2010, 17:28
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\(x^3  x = (xa)(xb)(xc)\) \(x(x^21)=(xa)(xb)(xc)\) \(x(x+1)(x1)=(xa)(xb)(xc)\) We know that \(a>b>c\). We also know that \(x+1>x>x1\). Now, rearrange the terms from highest to lowest on both sides. \((x+1)x(x1)=(xc)(xb)(xa)\) So you can see \(x+1 = xc\), \(x = xb\), \(x1=xa\). The easiest one is \(x = xb\). 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!



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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 Aug 2010, 03:23



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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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12 Aug 2013, 18:48
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^3x=(xa)(xb)(xc) and rearrange it like so:
(x^3x)/((xa)(xc))=xb
and then, since this equation should hold true for all numbers x, plug in x=1, which would give you:
0/((1a)(1c))=1b
The left side of the equation goes to 0, giving
0=1b, thus
b=1
Am I misunderstanding what is meant by "for all numbers x"? Thanks for your help!



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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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13 Aug 2013, 01:01
zx8009 wrote: 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^3x=(xa)(xb)(xc) and rearrange it like so:
(x^3x)/((xa)(xc))=xb
and then, since this equation should hold true for all numbers x, plug in x=1, which would give you:
0/((1a)(1c))=1b
The left side of the equation goes to 0, giving
0=1b, thus
b=1
Am I misunderstanding what is meant by "for all numbers x"? Thanks for your help! You cannot divide by (xa)(xc), since (xa)(xc) can be zero and division by zero is not allowed. Hope it's clear.
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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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jpr200012 wrote: If a, b, and c are constants, a > b > c, and x^3x=(xa)(xb)(xc) for all numbers x, what is the value of b?
A. 3 B. 1 C. 0 D. 1 E. 3 x^3x=(xa)(xb)(xc) or, x (x^2 1) = (xa)(xb)(xc) or, x (x+1)(x1) = (xa)(xb)(xc) or, (x1)(x0){x(1)} = (xa)(xb)(xc) Given, a > b > c Same, 1>0>1 Finally, 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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If a, b, and c are constants, a > b > c, and x^3  x = (x  a)(x  b) [#permalink]
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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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Option C: x^3x can be rewritten as:(x1)*x*(x+1) in ascending order. (x1)*x*(x+1)=(xa)*(xb)*(xc) [As a>b>c=>a<b<c => xa<xb<xc] Therefore,a=1 b=0 c=1



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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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22 Feb 2014, 05:03
SOLUTIONIf 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 As \(x^3x=(xa)(xb)(xc)\) is true for ALL \(xes\) than it must be true for \(x=0\), \(x=1\) and \(x=1\) too: \(x=0\) > \(x^3x=(xa)(xb)(xc)\) becomes: \(0=abc\), so one of the unknowns equals to zero; \(x=1\) > \(x^3x=(xa)(xb)(xc)\) becomes: \(0=(1a)(1b)(1c)\), so one of the unknowns equals to 1; \(x=1\) > \(x^3x=(xa)(xb)(xc)\) becomes: \(0=(1a)(1b)(1c)\), so one of the unknowns equals to 1. As \(a > b > c\) then \(a=1\), \(b=0\) and \(c=1\). Answer: C.
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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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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
\(x^3  x = (xa)(xb)(xc)\) \(x(x^2  1) = (xa)(xb)(xc)\)
Using \((ab)^2 = (ab)(a+b)\) algebra form, the expression becomes \(( x ) (x  1)(x + 1) = (xa)(xb)(xc)\)
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



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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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\(x^3  x = (x  a)(x  b)(x  c)\) \(x (x^2  1) = (x  a)(x  b)(x  c)\) (x + 0) (x+1) (x1) = (x  a)(x  b)(x  c) Given that a>b>c; so reorganising LHS of the equation (x1) (x + 0) (x+1) = (x  a)(x  b)(x  c) b = 0 Answer = C
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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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17 Jul 2014, 04:17
How do we know so is equal the numbers are 1,0,1. I know there consecutive but why not 2,3,4?



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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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17 Jul 2014, 04:43
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? Please check here: ifabandcareconstantsabcandx3xx167671.html#p1335360As x^3x=(xa)(xb)(xc) is true for ALL xes 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.
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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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17 Jul 2014, 17:52
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? Expanding \(x^3  x\) gives the values 1, 0, 1 besides x. Those are derived as shown in my above post \((x+2)(x+3)(x+4) \neq{x^3  x}\)
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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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23 Oct 2014, 20:58
jpr200012 wrote: If a, b, and c are constants, a > b > c, and x^3x=(xa)(xb)(xc) for all numbers x, what is the value of b?
A. 3 B. 1 C. 0 D. 1 E. 3 X^3  X = (XA) (XB) (XC) X (X^21) = (XA) (XB) (XC) X (X1) (X+1) = (XA) (XB) (XC) Since, A > B> C, Therefore, (XA) < (XB) < (XC) (X1) X (X+1) = (XA) (XB) (XC) (XA) = (X1)  A = 1 (XB) = X  B = 0 (XC) = (X+1)  C = 1 Therefore Answer is OA C



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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, 12:11
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^3x=(xa)(xb)(xc) is true for ALL xes 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 xes"  I really dont understand this.




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