In any triangle `A B C ,` prove that: `\ a^3sin(B-C)+b^3sin(C-A)+c^3sin(A-B)=0`

Explanation

To prove that \( a^3 \sin(B - C) + b^3 \sin(C - A) + c^3 \sin(A - B) = 0 \) in any triangle \( ABC \), we can use the sine rule and some trigonometric identities. Here’s a step-by-step solution:

### Step 1: Use the Sine Rule
According to the sine rule, we have:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k \]
where \( k \) is a constant. Thus, we can express \( a, b, c \) in terms of \( k \):
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