# Algebraic Manipulation Proof

1. Jan 28, 2008

### Frillth

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

I need to prove the following equation is true, given that a + b + c = 0:
a^2/(a^2 - b^2 - c^2) + b^2/(b^2 - a^2 - c^2) + c^2/(c^2 - a^2 - b^2) = 3/2

2. Relevant equations

a + b + c = 0

3. The attempt at a solution

I wasn't sure what to do at first, so I tried cross multiplying the equation and combining like terms. I came up with:

-a^6 - b^6 - c^6 + a^4b^2 + a^4c^2 + b^4a^2 + b^4c^2 + c^4a^2 + c^4b^2 + 6a^2b^2c^2 = 0

It looks like this should factor out nicely, but I can't seem to figure out how. Can anybody help me out?

Edit: Finally got it solved. This was way easier than I tried to make it.

Last edited: Jan 29, 2008
2. Jan 28, 2008

### PowerIso

That makes me dizzy. Just inspecting it (i'll work on it later) it might be a bit easier to note that a + b + c that there exist at least 1 negative value in there (assuming a, b, c are real). From there you could possibly prove by cases that will equal to 3/2.

3. Jan 28, 2008

### Frillth

Yeah, that is a much nicer way of writing what I had. Sorry about the plain text, I just haven't used latex before.

Edit: Somebody else posted here earlier, but they must have deleted their post. It was just a reworking of my problem in latex.

Last edited: Jan 28, 2008
4. Jan 29, 2008

### arildno

Just set, say, c=-(a+b) and plug it in wherever c occurs. Then simplify.

By the way, a+b+c=0 is an equation, whereas the other is an identity.

5. Jan 29, 2008

### Dick

Notice that if your quadratic identity is true for a,b and c then it's also true for ra, rb and rc for any nonzero constant r. So you could just choose, say a=1. So if you can prove it for a=1, c=-(1+b), then you've proved it for all a,b,c (at least the ones for which it is defined).