About halfway through a homework problem, I end up with a three-way identity:
$$\frac{uw}{v+w} = \frac{uv}{u+w} = \frac{vw}{u+v}$$
(I say I end up with..., but this is the method suggested by my professor.) What I have to do is show that $u=v=w=1$. I started by simplifying two of the expressions:
$$\frac{u}{u+w} = \frac{w}{u+v}$$
From this, we can conclude that $v = \frac{\left(u+w \right)w-u^2}{u}$, which can then be substituted into another of the above identities, say $\frac{uw}{v+w} = \frac{uv}{u+w}$.
Eventually, this leads to a fourth-degree polynomial equation:
$$ \begin {align} \ w^4+3w^3u - w^2 u^2 - 4w u^3 + u^4 &= 0 \\ \ (w - u)(w^3 + 4w^2 u + 3wu^2 - u^3) &= 0 \\ \end{align} $$
I factored the above by long division, simply because I knew that $u = w$ is supposed to be a solution. Beyond this point, though, I'm not sure what to do. Wolfram Alpha says the cubic is irreducible; that seems totally plausible if we're working in $\mathbb{Q}[u,w]$. But this is a third-degree polynomial: Shouldn't there be at least one linear factor in $\mathbb{R}[u,w]$?
I was able to substitute to get the equation $k^3 + 4k^2 + 3k - 1 = 0$, which I was able to rework into the depressed cubic $z^3 - \frac{7}{3}z - \frac{7}{27} = 0$. I tried using Bombelli's method, which gave me
$$z = \sqrt[3]{\frac{7}{54}+\frac{7}{54}i\sqrt{755}}+\sqrt[3]{\frac{7}{54}-\frac{7}{54}i\sqrt{755}}$$
I set the first term on the RHS to $p+qi$, cubing both sides and trying to match up the real parts (i.e., $p\left(p^2 - 3q^2 \right) = 7 / 54$), but that's as far as I got.
So my question: Is this a viable route toward the answer? If so, how might I proceed? Do I need to take a few steps back? Or should I go back to my first step and pick a totally different path? (Perhaps there's a way to keep the degree of the equation down by starting again at the beginning?) I realize the answers to all of these questions may be yes, and so, if they are, hopefully I can get suggestions for multiple methods: I'm really just interested in methodology here, I guess.
One final note: $u, v, w$ are the weights of three points using barycentric coordinates. (Hopefully I'm saying that right.) Therefore, each must be positive; that's the only restriction I'm aware of. Of the approximate roots of the depressed cubic — Wolfram gives the exact answers as complex, though the approximations have no imaginary part — only one is positive; therefore there's only one extra answer I'm concerned about eliminating — either by not letting it in in the first place or by being able to explain why it's not a valid answer. (I realize there's a lot here, but hopefully it's clear; please let me know if anything isn't.)
Making the assumption that $u,v,w$ are positive (last paragraph), we get that
$$ \frac{ uvw} { v^2 + vw } = \frac{ uvw } { w^2 + uw} = \frac {uvw}{u^2 + uv} ,$$
and hence that
$$ v^2 + vw = w^2 + wu = u^2 + uv .$$
Now, take the difference of 2 equations, we get that
$$ v^2 - w^2 = w (u-v), w^2 - u^2 = u(v-w), u^2 - v^2 = v(w-u). $$
Multiplying these 3 equations together, we get
$$ (v-w)(v+w)(w-u)(w+u)(u-v)(v+u) = (u-v)(v-w)(w-u)wuv \\ \Rightarrow (u-v)(v-w)(w-u) [ (v+w) ( w+u)(u+v) - uvw] = 0 . $$
Since $ (v+w) ( w+u)(u+v) \geq 2 \sqrt{vw} 2 \sqrt{wu} 2 \sqrt{uv} > uvw$, hence we must have $ (v-w)(w-u)(u-v) = 0$. It is easy to show that if $ v=w$, then we must have $w=u$ in the original equations.
Hence, $ u=v=w$ is the only set of solutions.