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March 17th, 2015, 05:59 AM   #1
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Factorization

Let a,b,c be numbers such that:
a²-ab=1
b²-bc=1
c²-ac=1
The value of abc(a+b+c) is equal to:
a)0
b)1
c)2
d)-1
e)-3

answer: d)
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March 17th, 2015, 07:06 PM   #2
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hello!?
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March 17th, 2015, 07:14 PM   #3
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Originally Posted by nandofab View Post
hello!?
Please don't bump threads like this. If you have made progress on a problem for which you have gotten no reply, feel free to post that progress, but we ask that you do not post for the sole reason to raise a thread's profile. This adds no value to a thread.

Many of us who volunteer help are not very inclined to help someone who posts a problem with no effort shown. You are more likely to garner replies if you post your work so far, or at least your thoughts on how you should begin.
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March 17th, 2015, 08:55 PM   #4
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MarkFL. I was 'him' last time. Hahaha.
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March 19th, 2015, 07:14 PM   #5
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Ok, I'm very sorry! From now on, I'll post what I tried to do. Thanks
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March 21st, 2015, 06:41 AM   #6
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Multiplying the first equation by (1 - 3bc) gives a² - ab - 3a²bc + 3ab²c = 1 - 3bc.
Similarly, b² - bc - 3ab²c + 3abc² = 1 - 3ac and c² - ac - 3abc² + 3a²bc = 1 - 3ab.
Adding these three results gives a² + b² + c² + 2ab + 2bc + 2ac = 3, i.e. (a + b + c)² = 3.

Multiplying the first equation by (bc² + c) gives a²bc² - ab²c² + a²c - abc = bc² + c.
Similarly, a²b²c - a²bc² + ab² - abc = a²c + a and ab²c² - a²b²c + bc² - abc = ab² + b.
Adding these three results gives -3abc = a + b + c.

Hence abc(a + b + c) = -(a + b + c)²/3 = -1.
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