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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 $(x+y)^3=x^3+y^3,$ you correctly surmise that $3xy^2+3x^2y=0,$ but just factorising this and dividing by 3 gives you $xy(x+y)=0.$ What does this tell you?
September 18th, 2009, 11:57 AM   #3
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 Originally Posted by mattpi It's a bit longwinded - if $(x+y)^3=x^3+y^3,$ you correctly surmise that $3xy^2+3x^2y=0,$ but just factorising this and dividing by 3 gives you $xy(x+y)=0.$ What does this tell you?
right.. I need to pay more attention, thanks

 September 19th, 2009, 01:12 AM #4 Global Moderator   Joined: Dec 2006 Posts: 20,935 Thanks: 2209 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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