Consider the following question
∣∣ ∣ ∣∣aa2bcbb2cacc2ab∣∣ ∣ ∣∣=(a−b)(b−c)(c−a)(ab+bc+ca)
Applying R2→R2−R1 R3→R3−R1
1abc∣∣ ∣ ∣∣a2a3abcb2b3bcac2c3cab∣∣ ∣ ∣∣=abcabc∣∣ ∣ ∣∣a2a31b2b31c2c31∣∣ ∣ ∣∣
∣∣ ∣ ∣∣a2a31b2b31c2c31∣∣ ∣ ∣∣=∣∣ ∣ ∣∣a2a31b2−a2b3−a30c2−a2c3−a30∣∣ ∣ ∣∣
=(b2−a2)(c3−a3)−(b3−a3)(c2−a2)
=(b−c)(b+c)(c−a)(c2+a2+ac)−(b2+a2+ab)(a+b)(c−a)(c+a)
=(c−a)(b−a)((b−c)(b+c)(c2+a2+ac)−(b2+a2+ab)(a+b)(c+a))
=(c−a)(b−a)(a−c)(ab+bc+ca)
Hence, this is the required answer.