# Algebra proof question

1. Apr 8, 2014

### smart_worker

2. Relevant equations
If a(y+z)=x, b(z + x)=y, c(x+y)=z, prove that x2/a(1-bc) = y2/b(1-ca) = z2/c(1-ab)

3. The attempt at a solution

ay+az=x,bz+bx=y,cx+cy=z.
i solved these equations and found x,y and z values,pluged them onto the proof equation.but it took me about half an hour for that.is there a quicker or easier way to approach these kind of questions?

2. Apr 8, 2014

### haruspex

Seeing the form of the desired result, I used the first equations to find expressions for a, b, c in terms of x, y, z. Then I substituted these in a(1-bc) and got x2(x+y+z)/((x+y)(y+z)(z+x)), and it's obvious from there.

3. Apr 9, 2014

### epenguin

I don't know in what level of course this question came - group theory or abstract algebra?

Whatever, I wonder if you are supposed to say something like: you have an equality* that is unaltered by moving its terms one step forward alphabetically in cyclic fashion:

a → b → c → a, x → y → z → x.

This must be true still of any equality you derive from the given ones.

Therefore there is no way x2/a(1 - bc) could not be equal to the last two.

?

*or, I think you can frame it, just a quantity, or something more abstract, x.

Last edited: Apr 9, 2014
4. Apr 9, 2014

### smart_worker

u have gone too deep,all the question i am asking are from higher algebra-hall & knight

5. Apr 9, 2014

### haruspex

Does that work? If the given equations were a=x, b=y, c=z, does it follow that a/y = b/z = c/x?
I think you first need to show one of the target equalities before you can apply cyclicity to derive the other.

6. Apr 24, 2014

### epenguin

Yes what I said is wrong. You do have to show one of the target equalities, the other is then automatically true, it is saying the same thing as the first with names changed.

I couldn't resist the thought that because we have a form f(a, b, c, x, y, z) = f(b, c, a, y, z, x) ( = f(c, a, b, z, x, y) ) then any other function g(a, b, c, x) = g(b, c, a, y) (I mean when the terms are connected by the f relation) . But it can't be generally true or else I could invent various different g of which some would give results contradicting others. It would make things like this so easy I wondered if there is a condition for it to be true that one could easily see which could save a lot of work, but this is probably a rubbish thought.

The solution expected from you is these are just three simultaneous linear equations from which you can eliminate z, say in three different ways to get

x/y = a(1 + c)/(1 - ac) = (1 - bc)/b(1 + c) = a(1 + b)/b(1 + a) .

You then get the result for x2/y2 multiplying the second expression above by the third.

You are meant to realise that the three apparently different expressions for x/y are the same thing as each other in virtue of a relation that must hold between x, y, z (independently of a, b, c)

ab + ac + bc + 2ab -1 = 0.

for the equations to have solutions (other than x = y = z = 0). You can get this condition from any pair of the above equations.

So you can only fix at will two of a, b, c then the third is dependent. After which the ratios x: y: z are fixed and you can decide at will only one of these three.

Last edited: Apr 25, 2014
7. Apr 28, 2014

### smart_worker

I solved the question in the way haruspex had mentioned but still I like your approach too.