In a triangle ABC, a, b, c are the sides of t | vedclass

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(A) Using the sine rule, $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$, so $a = k \sin A, b = k \sin B, c = k \sin C$. The expression i

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(A) Using the sine rule, $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$, so $a = k \sin A, b = k \sin B, c = k \sin C$. The expression is $\sum a^{3} \sin (B-C) = k^{3} \sum \sin^{3} A \sin (B-C)$. Since $A = 180^{\circ} - (B+C)$, $\sin A = \sin (B+C)$. Thus, the expression becomes $k^{3} \sum \sin^{2} A \sin (B+C) \sin (B-C)$. Using $\sin (B+C) \sin (B-C) = \sin^{2} B - \sin^{2} C$, we get: $k^{3} [\sin^{2} A (\sin^{2} B - \sin^{2} C) + \sin^{2} B (\sin^{2} C - \sin^{2} A) + \sin^{2} C (\sin^{2} A - \sin^{2} B)]$. Expanding this, we get $k^{3} [\sin^{2} A \sin^{2} B - \sin^{2} A \sin^{2} C + \sin^{2} B \sin^{2} C - \sin^{2} B \sin^{2} A + \sin^{2} C \sin^{2} A - \sin^{2} C \sin^{2} B] = 0$.

In a $\triangle ABC$, $a, b, c$ are the sides of the triangle opposite to the angles $A, B, C$ respectively. Then, the value of $a^{3} \sin (B-C) + b^{3} \sin (C-A) + c^{3} \sin (A-B)$ is equal to

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