$\underline{\textbf{Solution 1 }}$ \begin{align} & a^2(\sin(A-B)-\sin(A+B))=b^2(-\sin(A+B)-\sin(A-B)) \\ \implies & a^2\cos{A}\sin{B}=b^2\cos{B}\sin{A}\\ \implies & \sin^2{A}\cos{A}\sin{B}=\sin^2{B}\cos{B}\sin{A}\\ \implies & \sin{A}\sin{B}(\sin{A}\cos{A}-\sin{B}\cos{B})=0\\ \implies & \sin{A}\sin{B}(\sin{2A} - \sin{2B})=0 \end{align} Therefore $\sin 2A = \sin 2B \implies A=B$ or $2A+2B=\pi$. Hence, $\triangle{ABC}$ is either a right triangle or an isosceles triangle. $\underline{\textbf{Solution 2}}$ \begin{align*} & a^2(\sin(A-B)-\sin(A+B))=b^2(-\sin(A+B)-\sin(A-B)) \\ \implies & a^2\cos{A}\sin{B}=b^2\cos{B}\sin{A}\\ \implies & a^2b\dfrac{b^2 + c^2-a^2}{2bc} = b^2a\dfrac{a^2+c^2-b^2}{2ac}\\ \implies & a^2(b^2+c^2-a^2)=b^2(a^2+c^2-b^2)\\ \implies & (a^2-b^2)(a^2+b^2-c^2)=0 \end{align*} Hence, either $a=b$ or $a^2 + b^2=c^2$. This means that the triangle is either isosceles or right.