# Zeros proof math

1. Sep 4, 2010

### jgens

1. The problem statement, all variables and given/known data

Prove that if x3+x2y+xy2+y3 = 0, then x = y = 0 or x = -y.

2. Relevant equations

N/A

3. The attempt at a solution

Assume that x3+x2y+xy2+y3 = 0, in which case, it follows that x3+y3 = -(x2y+xy2) or (x+y)(x2-xy+y2) = -xy(x+y). Equality clearly holds if x+y = 0. Now, suppose that x+y =/= 0, and divide through by x+y. This leaves the equality x2-xy+y2 = -xy or x2+y2 = 0, which can only happen if x = y = 0. Therefore, if x3+x2y+xy2+y3 = 0, then x = -y or x = y = 0.

Does this 'proof' work?

2. Sep 4, 2010

### Mentallic

Re: Zeros

Yes, of course it does. Your proof is sound and logical.

3. Sep 4, 2010

### jgens

Re: Zeros

Thanks! I've been second guessing myself a lot lately, so it's always good to get confirmation that I've done something correctly.

4. Sep 4, 2010

### Mentallic

Re: Zeros

Yeah, I know the feeling

5. Sep 5, 2010

### HallsofIvy

Re: Zeros

Excellent! I would have done it a slightly different way: it appeared to me, looking at the first two terms, that $x^2+ x^2y= x^2(x+ y)$ and then that $xy^2+ y^3= y^2(x+ y)$. Since (x+ y) appears in both of those we can factor it out and have $x^3+ x^2y+ xy^2+ y^3= (x^2+ y^2)(x+ y)$.

That will be 0 only if one or the other of those factors is 0. $x^2+ y^2= 0$ only if x= y= 0 and x+ y= 0 only if y= -x.

6. Sep 5, 2010

### jgens

Re: Zeros

Thanks Halls! I like the way that your proof works out much better. It looks a lot more like what I typically see.

7. Sep 6, 2010

### HallsofIvy

Re: Zeros

On the other hand, your proof is much better for you than mine because it is yours!