Question 5(v) Show that: (a-b)(a+b)+(b-c)(b+c)+(c-a)(c+a)=0
L.H.S = (a−b)(a+b)+(b−c)(b+c)+(c−a)(c+a) =a2−b2+b2−c2+c2−a2 [Using identity (a−b)(a+b)=a2−b2] =0. =R.H.S