System of equations with three variables Characterize all triples $(a,b,c)$ of positive real numbers such that
$$ a^2-ab+bc = b^2-bc+ca = c^2-ca+ab. $$
This is the equality case of the so-called Vasc inequality. I think the answer is that $a=b=c$ or $a:b:c = \sin^2(4\pi/7) : \sin^2(2\pi/7) : \sin^2(\pi/7)$ and cyclic equality cases. I'm mostly interested if there is a way to derive this reasonably, other than just magically guessing the solutions and then doing degree-counting. This was left as a "good exercise" in Mildorf that has bothered me for years.
 A: Too long for comment 
Using $\sin(\pi-A)=+\sin A,\cos(\pi-B)=-\cos B,\sin2C=2\sin C\cos C$
$$\dfrac{\sin^2\dfrac{2\pi}7}{\sin^2\dfrac{\pi}7}=4\cos^2\dfrac\pi7=2+2\cos\dfrac{2\pi}7$$
$$\dfrac{\sin^2\dfrac{\pi}7}{\sin^2\dfrac{3\pi}7}=\dfrac{\sin^2\dfrac{6\pi}7}{\sin^2\dfrac{3\pi}7}=4\cos^2\dfrac{3\pi}7=2+2\cos\dfrac{6\pi}7$$
$$\dfrac{\sin^2\dfrac{3\pi}7}{\sin^2\dfrac{2\pi}7}=\dfrac{\sin^2\dfrac{4\pi}7}{\sin^2\dfrac{2\pi}7}=4\cos^2\dfrac{2\pi}7=2+2\cos\dfrac{4\pi}7$$
From How to solve $8t^3-4t^2-4t+1=0$ and factor $z^7-1$ into linear and quadratic factors and prove that $ \cos(\pi/7) \cdot\cos(2\pi/7) \cdot\cos(3\pi/7)=1/8$,
the roots of $$x^3+x^2-2x-1=0$$ are $2\cos\dfrac{2\pi}7,2\cos\dfrac{4\pi}7,2\cos\dfrac{6\pi}7$
Replace $x$ with $y-2$ to find $$y^3-5y^2+6y-1=0$$
A: If $c=0$ then $a^2-ab=b^2=ab$, which gives $a=b=c=0.$
Let $abc\neq0$ and $a=xb$.
Thus, from the first equation we obtain:
$$a^2-ab-b^2=(a-2b)c.$$
If $a=2b$ so $a=b=c=0$, which is impossible here.
Thus, $c=\frac{a^2-ab-b^2}{a-2b}$ and from the second equation we obtain:
$$b^2+\frac{(a-b)(a^2-ab-b^2)}{a-2b}=\frac{(a^2-ab-b^2)^2}{(a-2b)^2}-\frac{(a^2-ab-b^2)a}{a-2b}+ab$$ or
$$1+\frac{(x-1)(x^2-x-1)}{x-2}=\frac{(x^2-x-1)^2}{(x-2)^2}-\frac{(x^2-x-1)x}{x-2}+x$$ or
$$(x-1)(x^3-5x^2+6x-1)=0,$$ which gives $x=1$ and $a=b=c$ or
$$x^3-5x^2+6x-1=0.$$
Now, easy to show that $\frac{\sin^2\frac{2\pi}{7}}{\sin^2\frac{\pi}{7}}$, $\frac{\sin^2\frac{\pi}{7}}{\sin^2\frac{3\pi}{7}}$ and $\frac{\sin^2\frac{3\pi}{7}}{\sin^2\frac{2\pi}{7}}$ they are roots of the last equation.
For example:
$$\left(\frac{\sin^2\frac{2\pi}{7}}{\sin^2\frac{\pi}{7}}\right)^3-5\left(\frac{\sin^2\frac{2\pi}{7}}{\sin^2\frac{\pi}{7}}\right)^2+6\cdot\frac{\sin^2\frac{2\pi}{7}}{\sin^2\frac{\pi}{7}}-1=$$
$$=\left(4\cos^2\frac{\pi}{7}\right)^3-5\left(4\cos^2\frac{\pi}{7}\right)^2+6\left(4\cos^2\frac{\pi}{7}\right)-1=$$
$$=\left(2+2\cos\frac{2\pi}{7}\right)^3-5\left(2+2\cos\frac{2\pi}{7}\right)^2+6\left(2+2\cos\frac{2\pi}{7}\right)-1=$$
$$=8\cos^3\frac{2\pi}{7}+4\cos^2\frac{2\pi}{7}-4\cos\frac{2\pi}{7}-1=$$
$$=2\left(4\cos^3\frac{2\pi}{7}-3\cos\frac{2\pi}{7}\right)+6\cos\frac{2\pi}{7}+2+2\cos\frac{4\pi}{7}-4\cos\frac{2\pi}{7}-1=$$
$$=2\cos\frac{2\pi}{7}+2\cos\frac{4\pi}{7}+2\cos\frac{6\pi}{7}+1=$$
$$=\frac{2\sin\frac{\pi}{7}\cos\frac{2\pi}{7}+2\sin\frac{\pi}{7}\cos\frac{4\pi}{7}+2\sin\frac{\pi}{7}\cos\frac{6\pi}{7}}{\sin\frac{\pi}{7}}+1=$$
$$=\frac{\sin\frac{3\pi}{7}-\sin\frac{\pi}{7}+\sin\frac{5\pi}{7}-\sin\frac{3\pi}{7}+\sin\frac{7\pi}{7}-\sin\frac{5\pi}{7}}{\sin\frac{\pi}{7}}+1=0.$$
Since we have no another roots, we are done!
A: Suppose $a=b$, then $bc=b^2=c^2-cb+b^2$, those $a=b=c$. Which includes cyclic equality cases as well.
Now lets assume WLOG $a>b>c$. Show that in this case above equations don't hold simultaneously. 
