Question
$\frac{\text{a}^2\sin(\text{B}-\text{C})}{\sin\text{A}}+\frac{\text{b}^2\sin(\text{C}-\text{A})}{\sin\text{B}}+\frac{\text{c}^2\sin(\text{A}-\text{B})}{\sin\text{C}}=0$
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Answer
$\frac{\text{a}^2\sin(\text{B}-\text{C})}{\sin\text{A}}+\frac{\text{b}^2\sin(\text{C}-\text{A})}{\sin\text{B}}+\frac{\text{c}^2\sin(\text{A}-\text{B})}{\sin\text{C}}=0$
$\frac{\text{a}}{\sin\text{A}}=\frac{\text{b}}{\text{b}\sin\text{C}}=\frac{\text{c}}{\sin\text{C}}=\text{k}$
$\text{LHS}=\frac{\text{a}^2\sin(\text{B}-\text{C})}{\sin\text{A}}+\frac{\text{b}^2\sin(\text{C}-\text{A})}{\sin\text{B}}+\frac{\text{c}^2\sin(\text{A}-\text{B})}{\sin\text{C}}$
$=\text{ak}\sin(\text{B}-\text{C})+\text{bk}\sin(\text{C}-\text{A})+\text{ck}\sin(\text{A}-\text{B})$
$=\sin\text{A}\sin(\text{B}-\text{C})+\sin\text{B}\sin(\text{C}-\text{A})+\sin(\text{A}-\text{B})$
$=\sin(\pi-(\text{B + C}))\sin(\text{B}-\text{C})+\sin(\pi-(\text{C + A}))\\\sin(\text{C}-\text{A})+\sin(\pi-(\text{A + B}))\sin(\text{A}-\text{B})$
$=\sin(\text{B + C})\sin(\text{B}-\text{C})+\sin(\text{C + A})\\\sin(\text{C} - \text{A})+\sin(\text{A + B})\sin(\text{A}-\text{B})$
$=\sin^2\text{B}-\sin^2\text{C}+\sin^2\text{C}-\sin^2\text{A}+\sin^2\text{A}-\sin^2\text{B}=0=\text{RHS}$
$\frac{\text{a}}{\sin\text{A}}=\frac{\text{b}}{\text{b}\sin\text{C}}=\frac{\text{c}}{\sin\text{C}}=\text{k}$
$\text{LHS}=\frac{\text{a}^2\sin(\text{B}-\text{C})}{\sin\text{A}}+\frac{\text{b}^2\sin(\text{C}-\text{A})}{\sin\text{B}}+\frac{\text{c}^2\sin(\text{A}-\text{B})}{\sin\text{C}}$
$=\text{ak}\sin(\text{B}-\text{C})+\text{bk}\sin(\text{C}-\text{A})+\text{ck}\sin(\text{A}-\text{B})$
$=\sin\text{A}\sin(\text{B}-\text{C})+\sin\text{B}\sin(\text{C}-\text{A})+\sin(\text{A}-\text{B})$
$=\sin(\pi-(\text{B + C}))\sin(\text{B}-\text{C})+\sin(\pi-(\text{C + A}))\\\sin(\text{C}-\text{A})+\sin(\pi-(\text{A + B}))\sin(\text{A}-\text{B})$
$=\sin(\text{B + C})\sin(\text{B}-\text{C})+\sin(\text{C + A})\\\sin(\text{C} - \text{A})+\sin(\text{A + B})\sin(\text{A}-\text{B})$
$=\sin^2\text{B}-\sin^2\text{C}+\sin^2\text{C}-\sin^2\text{A}+\sin^2\text{A}-\sin^2\text{B}=0=\text{RHS}$
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