If any triangle `A B C` , that: `(asin(B-C))/(b^2-c^2)=(bsin(C-A))/(c^2-a^2)=(csin(A-B))/(a^2-b^2)`

Explanation

To prove the equality

\[ \frac{a \sin(B - C)}{b^2 - c^2} = \frac{b \sin(C - A)}{c^2 - a^2} = \frac{c \sin(A - B)}{a^2 - b^2} \]

for any triangle \( ABC \), we can use the Sine Rule and some algebraic manipulation. Here’s a step-by-step solution:

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