<s>[INST] 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.
 [/INST] 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 $$y^3-5y^2+6y-1=0$$
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