March 17th, 2015, 05:59 AM  #1 
Newbie Joined: Feb 2015 From: Brazil Posts: 23 Thanks: 0  Factorization
Let a,b,c be numbers such that: aÂ²ab=1 bÂ²bc=1 cÂ²ac=1 The value of abc(a+b+c) is equal to: a)0 b)1 c)2 d)1 e)3 answer: d) 
March 17th, 2015, 07:06 PM  #2 
Newbie Joined: Feb 2015 From: Brazil Posts: 23 Thanks: 0 
hello!?

March 17th, 2015, 07:14 PM  #3 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,131 Thanks: 440 Math Focus: Calculus/ODEs  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. 
March 17th, 2015, 08:55 PM  #4 
Senior Member Joined: Sep 2013 From: Earth Posts: 821 Thanks: 35 
MarkFL. I was 'him' last time. Hahaha.

March 19th, 2015, 07:14 PM  #5 
Newbie Joined: Feb 2015 From: Brazil Posts: 23 Thanks: 0 
Ok, I'm very sorry! From now on, I'll post what I tried to do. Thanks

March 21st, 2015, 06:41 AM  #6 
Global Moderator Joined: Dec 2006 Posts: 16,377 Thanks: 1174 
Multiplying the first equation by (1  3bc) gives aÂ²  ab  3aÂ²bc + 3abÂ²c = 1  3bc. Similarly, bÂ²  bc  3abÂ²c + 3abcÂ² = 1  3ac and cÂ²  ac  3abcÂ² + 3aÂ²bc = 1  3ab. Adding these three results gives aÂ² + bÂ² + cÂ² + 2ab + 2bc + 2ac = 3, i.e. (a + b + c)Â² = 3. Multiplying the first equation by (bcÂ² + c) gives aÂ²bcÂ²  abÂ²cÂ² + aÂ²c  abc = bcÂ² + c. Similarly, aÂ²bÂ²c  aÂ²bcÂ² + abÂ²  abc = aÂ²c + a and abÂ²cÂ²  aÂ²bÂ²c + bcÂ²  abc = abÂ² + b. Adding these three results gives 3abc = a + b + c. Hence abc(a + b + c) = (a + b + c)Â²/3 = 1. 

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