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\frac{a(a^2-bc) + b(b^2-ac) + c(c^2-ab)}{(a+b+c)(a^2-bc + b^2-ac + c^2-ab)}
\frac{a^3+b^3+c^3-3abc}{(a+b+c)(a^2+b^2+c^2-ab -bc -ca)}
\frac{a^3+b^3+c^3-3abc}{a^3+b^3+c^3-3abc}=1
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