Let A, B, C be three angles such that sin A + | vedclass

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(C) Given $\sin A + \sin B + \sin C = 0$.We know the identity $\sin 3	heta = 3\sin 	heta - 4\sin^3 	heta$.Thus,$\sin 3A + \sin 3B + \sin 3C = 3(\si

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$-\frac{1}{12}$

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(C) Given $\sin A + \sin B + \sin C = 0$.We know the identity $\sin 3	heta = 3\sin 	heta - 4\sin^3 	heta$.Thus,$\sin 3A + \sin 3B + \sin 3C = 3(\sin A + \sin B + \sin C) - 4(\sin^3 A + \sin^3 B + \sin^3 C)$.Since $\sin A + \sin B + \sin C = 0$,the expression simplifies to $-4(\sin^3 A + \sin^3 B + \sin^3 C)$.Using the identity: if $x+y+z=0$,then $x^3+y^3+z^3 = 3xyz$.Here,$\sin^3 A + \sin^3 B + \sin^3 C = 3 \sin A \sin B \sin C$.Substituting this into the denominator: $\sin 3A + \sin 3B + \sin 3C = -4(3 \sin A \sin B \sin C) = -12 \sin A \sin B \sin C$.Therefore,the ratio is $\frac{\sin A \sin B \sin C}{-12 \sin A \sin B \sin C} = -\frac{1}{12}$.

Let $A, B, C$ be three angles such that $\sin A + \sin B + \sin C = 0$. Then,$\frac{\sin A \sin B \sin C}{\sin 3A + \sin 3B + \sin 3C}$ (wherever defined) is equal to:

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