What is (a-b)²/((b-c)(c-a)) + (b-c)²/((c-a)(a-b)) + (c-a)²/((a-b)(b-c)) - 3 equal to, where a ≠ b ≠ c?

Using the well-known identity: if x + y + z = 0, then x³ + y³ + z³ = 3xyz. Let x = a-b, y = b-c, z = c-a, then x + y + z = 0, so x³ + y³ + z³ = 3xyz = 3(a-b)(b-c)(c-a). The given expression equals [(a-b)³ + (b-c)³ + (c-a)³]/[(a-b)(b-c)(c-a)] - 3 = 3(a-b)(b-c)(c-a)/[(a-b)(b-c)(c-a)] - 3 = 3 - 3 = 0.