Dado un triángulo \(ABC\) con \(BC=a\), \(CA=b\), \(AB=c\), \(\angle BAC = \alpha\), \(\angle CBA = \beta\), \(\angle ACB = \gamma\). Demuestra que \[ a \sin(\beta-\gamma) + b \sin(\gamma-\alpha) + c\sin(\alpha-\beta) = 0. \]