A(a−b)(b−c)(c−a)B(a+b)(b+c)(c+a)C3(a+b)(b+c)(c+a)D(a+b+c)3 Step-by-Step SolutionEach triple sums to zero, so numerator =3(a2−b2)(b2−c2)(c2−a2) and denominator =3(a−b)(b−c)(c−a). Cancelling leaves (a+b)(b+c)(c+a).Answer: B) (a+b)(b+c)(c+a)