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If abc is not equal to 0, what is the value of (a^3 + b^3 +

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If abc is not equal to 0, what is the value of (a^3 + b^3 + [#permalink] New post 16 Jul 2003, 01:16
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A
B
C
D
E

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(N/A)

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25% (01:06) correct 75% (01:19) wrong based on 2 sessions
If abc is not equal to 0, what is the value of (a^3 + b^3 + c^3)/(abc)?

(1). |a|=1, |b|=2, |c|=3
(2). a + b + c = 0

(A) Statement (1) ALONE is sufficient to answer the question, but statement (2) alone is not.
(B) Statement (2) ALONE is sufficient to answer the question, but statement (1) alone is not.
(C) Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient to answer the question.
(E) Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question.
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 [#permalink] New post 16 Jul 2003, 01:48
all the numbers are nonzero ones.

(1) is obviously not enough
consider a=1, b=2, c=3 and a=1, b=2, c=-3

(2) of no use alone
consider
a=10, b=-9, c=-1
a=1, b=1, c=-2

combine: the only possible combination for a+b+c=0 is a=1, b=2, c=-3

Thus, I vote for C.
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 [#permalink] New post 16 Jul 2003, 01:57
The answer is C.
1 alone is not enough. 2 alone is not enough, either.
1 and 2 taken together are possible only when
a) a=1, b=2, c=-3 or
b) a=-1, b=-2, c=3

but the results in both cases are the same
a) (1+8-27)/-6
b) (-1-8+27)/6 = (1+8-27)/-6
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 [#permalink] New post 16 Jul 2003, 02:17
terrific!
I'd vote for E. Cuz really there are two different solutions when we consider C. But I never calculate till end!!!
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 [#permalink] New post 16 Jul 2003, 02:27
C for me too...

Spilt up the equation to this form,

b^2+c^2/bc + a^2+c^2/ac + a^2+b^2/ab +6.
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 [#permalink] New post 16 Jul 2003, 02:33
Any more tries?

:twisted:
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 [#permalink] New post 16 Jul 2003, 03:04
try 1, 2, 3 = (1+8+27)/6=36/6=6
try 1, 2,-3 = (1+8-27)/(-6)=18/6=3

clearly (1) is not sufficient
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 [#permalink] New post 16 Jul 2003, 04:06
stolyar wrote:
try 1, 2, 3 = (1+8+27)/6=36/6=6
try 1, 2,-3 = (1+8-27)/(-6)=18/6=3

clearly (1) is not sufficient


Can't challenge a counterexample. Okay, we eliminate A and D. What now?
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 [#permalink] New post 16 Jul 2003, 04:31
It is B, unfortunately.
But why?
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 [#permalink] New post 16 Jul 2003, 05:42
OKies got my mistake while working through it.

I just thought on reducing the values A^3+b^3+C^3 and the following formula striked me.

a^3 + b^3 + c^3 = (a+b+c)(a^2-ab-ac-bc+b^2+c^2) + 3abc

So Considering that if we use B, we have the answer as 3.

Hope that's right.. :)
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 [#permalink] New post 16 Jul 2003, 05:55
I guess if such type of problem appaear on GMAT, all what is required is nice clean trick striking at the correct moment othwerwise, one would land up wasting 4 to 5 minutes solving it for actual.
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 [#permalink] New post 16 Jul 2003, 05:55
evensflow wrote:
OKies got my mistake while working through it.

I just thought on reducing the values A^3+b^3+C^3 and the following formula striked me.

a^3 + b^3 + c^3 = (a+b+c)(a^2-ab-ac-bc+b^2+c^2) + 3abc

So Considering that if we use B, we have the answer as 3.

Hope that's right.. :)


What is the formula that you use for a sum of cubes?
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 [#permalink] New post 16 Jul 2003, 06:32
My idea behind the thought was...

a^3 + b^3 = (a + b)(a^2 -ab + b^2)

Work on the lines the formula for sum of 3 cubes is as given in the answer earlier.

HTH
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 [#permalink] New post 16 Jul 2003, 08:12
My Ans is E:

From 2: It can get many solutions. USELESS.

From (1) & (2) togather, we get two sets of numbers.
Both sets give different answers.

Thus, answer is E
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 [#permalink] New post 16 Jul 2003, 09:19
The correct answer is B.

(1) can be shown to be insufficient with a quick example as most of you have shown.

Many of you also said the (2) does not tell you anything. Did you even try any numbers? The answer will always be the same. Although that does not PROVE it, it would certainly give you a clue that there is a way to simplify the sum of cubes. (Frankly, on the real test, if I tried 4 different sets of simply numbers and they all worked, I would assume it is true). You can prove it as follows:

In (2), we are given a + b + c = 0. This means c = -(a + b). Make this substitution in the sum of cubes, expand what used to be the c^3 term, then simplify. You will get -(a+b)*3*a*b. Now substitute c for the -(a+b) term and you will get 3*a*b*c.

If a + b + c = 0, then (a^3 + b^3 + c^3)/(abc) = 3abc/abc = 3 for ANY a, b, or c not equal to 0, hence B is the correct answer.
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  [#permalink] 16 Jul 2003, 09:19
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