Simplify: [(a²−b²)³ − (b²−c²)³ + (c²−a²)³] / [(a−b)³ + (b−c)³ + (c−a)³]

Using the identity: if x+y+z=0, then x³+y³+z³=3xyz. Let x=a²−b², y=b²−c², z=c²−a²; x+y+z=0. Numerator = 3(a²−b²)(b²−c²)(c²−a²). Denominator with a−b, b−c, c−a also sums to 0, so = 3(a−b)(b−c)(c−a). Simplifying: 3(a+b)(b+c)(c+a)(a−b)(b−c)(c−a) / 3(a−b)(b−c)(c−a) = (a+b)(b+c)(c+a). This matches option A rearranged as (a−b)(b+c)(c+a) is not quite right; the answer is (a+b)(b+c)(c+a).