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If abc 0, what is the value of a^3 + b^3 + c^3/abc ? (1)

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If abc 0, what is the value of a^3 + b^3 + c^3/abc ? (1) [#permalink] New post 22 Mar 2011, 10:05
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If abc ≠ 0, what is the value of \(\frac{a^3 + b^3 + c^3}{abc}\) ?

(1) \(|a|=1, |b|=2, |c|=3\)
(2) \(a + b + c = 0\)
[Reveal] Spoiler: OA

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Re: Inequalities and modules DS [#permalink] New post 22 Mar 2011, 10:28
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Statement 1:

|a|=1, |b|=2, |c|=3

So a, b, c can be positive or negative. Thus, the value of the expression cannot be uniquely determined.

NS

Statement 2:

a + b + c = 0

(a + b + c)^3 = a^3 + b^3 + c^3 + (3ab + 3bc + 3ca) (a + b + c) – 3abc

Substitute for a + b + c = 0 in the above equation:

0 = a^3 + b^3 + c^3 - 3abc

=> 3abc = a^3 + b^3 + c^3

=> (a^3 + b^3 + c^3)/abc = 3

Sufficient
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Re: Inequalities and modules DS [#permalink] New post 22 Mar 2011, 10:29
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since abc ≠ 0 , we know that either all numbers can be +ve or any two can be -ve

Option A does not help to solve the problem since we can not make out that which numbers can be -ve and value changes with every trial and error method. Therefore, A can not be the answer.

Option B tells us that A+B+C=0, therefore we can use the formula that

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

and since we know that a+b+c = 0 then \(a^3 + b^3 + c^3 = 3abc\)

And thus putting "3abc" at the place of \(a^3 + b^3 + c^3\), will give us the answer.
Therefore answer is "B"

Only thing that since I know the formula thats why I could solve this problem within seconds ..... how to solve if we do not know the formula .... please someone explain.
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Re: Inequalities and modules DS [#permalink] New post 22 Mar 2011, 10:30
NOTE: a^3 + b^3 + c^3 = 3abc if a + b + c = 0
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Re: Inequalities and modules DS [#permalink] New post 22 Mar 2011, 10:31
gmatpapa wrote:
If abc ≠ 0, what is the value of \(\frac{a^3 + b^3 + c^3}{abc}\) ?

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


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

Using statement 2 & the above formula:
\(\frac{a^3 + b^3 + c^3}{abc} = \frac{(a+b+c)(a^2+b^2+c^2-ab-bc-ac)}{abc}+\frac{3abc}{abc} = 0+3=3\)
Sufficient.

Ans: "B"
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Re: Inequalities and modules DS [#permalink] New post 22 Mar 2011, 10:32
By the way, are we expected to know these equations for GMAT? :o
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Re: Inequalities and modules DS [#permalink] New post 22 Mar 2011, 17:57
1. Not sufficient as there more than one possible combination for a, b and c that satisfies the given equation.

2. Sufficient
by solving the given equation we can see that
(a+b+c)^3 = a3+b3+c3-3abc

=> a3+b3+c3 = 3abc , enough to answer the question.

Answer is B.
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Re: Inequalities and modules DS [#permalink] New post 22 Mar 2011, 18:31
To add some more, this has been posted earlier, please take a look at the solution step suggesting c = -(a+b).

ds-1543.html
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Re: Inequalities and modules DS [#permalink] New post 23 Mar 2011, 06:23
I seriously don't know how to use this in real life. I am not good in algebra. Not my style :-(
Re: Inequalities and modules DS   [#permalink] 23 Mar 2011, 06:23
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