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September 18th, 2009, 11:41 AM  #1 
Senior Member Joined: Apr 2009 Posts: 201 Thanks: 0  is this a valid proof?
Hello, I was hoping that you guyscould check my work: show that (x+y)^3 = x^3 + y^3 only if x = 0, y = 0 or x = y. So, x^3 + 3x^2y + 3xy^2 + y^3 = x^3 + y^3 > 3xy^2 + 3x^2y = 0. The first case is trivial, if x and y are both zero then the equation equals to zero (I'm not sure if I should prove that, it seems a bit tedious) then, if x does not equal to zero, 3xy^2 =  3x^2y > 3y^2 x * x^1 = 3x^2 * x^1 y > 3y^2 =  3yx (by inverses and commutativity) So, if a * b = a * c = 0, either a is zero or b =c  if 3y*y = 3y*x then either 3y = 0 or y = x. If 3y = 0 then y = 0 If y does not equal to zero, repeat the same method revolving 3x and x = y 3xy^2 =  3x^2y > 3y^2x * y^1 = 3x^2 y * y^1 > 3x^2 =  3yx So, if a * b = a * c = 0, either a is zero or b =c  if 3y*y = 3y*x then either 3x = 0 or x = y. If 3x = 0 then x = 0. I'm not sure what I should include in my proofs, some things just seem to be so tedious and unnecessary (like proving that (ab) = (a)(b) or (a)(b) every time, or a*0 = 0). thanks 
September 18th, 2009, 11:50 AM  #2 
Senior Member Joined: May 2008 From: York, UK Posts: 1,300 Thanks: 0 
It's a bit longwinded  if you correctly surmise that but just factorising this and dividing by 3 gives you What does this tell you?

September 18th, 2009, 11:57 AM  #3  
Senior Member Joined: Apr 2009 Posts: 201 Thanks: 0  Re: Quote:
 
September 19th, 2009, 01:12 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 20,464 Thanks: 2038 
The answer follows immediately without having to divide by three first.

September 19th, 2009, 05:05 AM  #5 
Senior Member Joined: May 2008 From: York, UK Posts: 1,300 Thanks: 0 
Well, of course... it's just neater without the 3 

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