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 January 2nd, 2013, 12:32 AM #1 Newbie   Joined: Dec 2012 Posts: 13 Thanks: 0 Different complex numbers equality We have a,b,c different complex numbers so (a+b)^3 = (b+c)^3 = (c+a)^3 Show that a^3 = b^3 = c^3 From the first equality I reached a^3 - c^3 + 3b(a-c)(a+b+c) = 0 How a is different from c => a-c is different from 0 How do I show that a^3 - c^3 = 0?
 January 2nd, 2013, 04:07 AM #2 Newbie   Joined: Dec 2012 Posts: 13 Thanks: 0 Re: Different complex numbers equality No one?
 January 2nd, 2013, 08:52 AM #3 Senior Member   Joined: Mar 2012 From: Belgium Posts: 654 Thanks: 11 Re: Different complex numbers equality this one looks very easy to me :/ $(a+b)^3= (b+c)^3 = (c+a)^3$ so $(a+b)= (b+c) = (c+a)$ $(a+b)= (b+c)$ => $a=c$(1) $(b+c)= (c+a)$ => $b=a$(2) (1)+(2) ==> $b=c$ $a=b=c$ so $a^3=b^3=c^3$
 January 2nd, 2013, 10:35 AM #4 Newbie   Joined: Dec 2012 Posts: 13 Thanks: 0 Re: Different complex numbers equality Sorry man, it is not that simple. This problem is from the Math Gazette - high school section. As I said, I first tried to calculate (a+b)^3 = (b+c)^3 then (b+c)^3 = (a+c)^3 then (a+b)^3 = (a+c)^3 but I'm struck as I said in my first post.
January 2nd, 2013, 11:01 AM   #5
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Re: Different complex numbers equality

Quote:
 Originally Posted by gelatine1 $a=b=c$
a,b and c are different complex numbers .

January 3rd, 2013, 12:27 AM   #6
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Re: Different complex numbers equality

Quote:
Originally Posted by zaidalyafey
Quote:
 Originally Posted by gelatine1 $a=b=c$
a,b and c are different complex numbers .
o right thought these were ment to be normal variables

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