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September 18th, 2009, 11:41 AM   #1
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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
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September 18th, 2009, 11:50 AM   #2
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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?
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September 18th, 2009, 11:57 AM   #3
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Re:

Quote:
Originally Posted by mattpi
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?
right.. I need to pay more attention, thanks
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September 19th, 2009, 01:12 AM   #4
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The answer follows immediately without having to divide by three first.
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September 19th, 2009, 05:05 AM   #5
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Well, of course... it's just neater without the 3
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