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a+b)^2 * (b-c)] / >= 0 ? 1- a>b 2- b>c

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a+b)^2 * (b-c)] / >= 0 ? 1- a>b 2- b>c [#permalink] New post 06 Nov 2004, 18:08
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A
B
C
D
E

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[(a+b)^2 * (b-c)] / [(a-b)^3 * (b-c)^3] >= 0 ?

1- a>b
2- b>c
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 [#permalink] New post 07 Nov 2004, 10:38
I got A.

(a+b)^2 (b-C) / (a-b)^3 (b-c)^3 =
= Positive / (a-b)^3.

So knowing that a>b is enough to know that the whole thing is positive.
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 [#permalink] New post 07 Nov 2004, 11:08
I got it as C

from 1) we dont get any info about the relation betn b and c -- hence insufficient.

2) no info about a and b--- insufficient.

combining, we get a>b>c.
plug in a=-1, b=-2, c=-3
we get LHS >0.
hence C.
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 [#permalink] New post 07 Nov 2004, 11:43
I go with C. We need to know that b is not equal to C or a is not equal to b. If a=b or b=c, then the expression will be undefined.

So, C.
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 [#permalink] New post 07 Nov 2004, 14:01
OA on this one is A. It is also what I had but because of the reason mentioned by venksune, I believe that we need to know that b is different from c otherwise the expression is undefined. If the question clearly stated at first that a, b, and c are different numbers, then A would have been good because all we care for, as Dookie mentioned, is the expression (a-b)^3 which could give a negative/positive answer. In retrospect, C seems to be the answer here. By the way, this is from Kaplan800.
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  [#permalink] 07 Nov 2004, 14:01
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