By using properties of determinants, in Exercises 8 to 14, show that:

By using properties of determinants, in Exercises 8 to 14, show that:

=> (b-a)(c-a)(c² –b² + ac –ab  +a² –a²)
=> (b-a)(c-a){(c –b)(c+b) + a(c –b)}

=>(b-a)(c-a)(c-b)(a+b+c)

=> (a-b)(b-c)(c-a)(a+b+c)

Hence proved