`|[1,a, bc] ,[1, b, ca], [1, c, ab]|` =`(a-b)(b-c)(c-a)`

|[1, 1, 1], [a, b, c], [bc, ca, ab]| = (a-b)(b-c)(c-a)

|[1,1,1],[a,b,c],[bc,ca,ab]|=(a-b)(b-c)(c-a)

Using properties of determinant show that: det[[1,a,-bc1,b,-ca1,c,-ab]]=(a-b)(b-c)(c-a)det[[1,b,-ca1,c,-ab]]=(a-b)(b-c)(c-a)

Prove that |[1,a,bc] , [1,b,ca], [1,c,ab]|=|[1,a,a^2] , [1,b,b^2] , [1,c,c^2]|

|[bc, b+c, 1], [ca, c+a, 1], [ab, a+b, 1]|=?

Prove the following : |{:(1,1,1),(a,b,c),(bc,ca,ab):}|=(a-b)(b-c)(c-a)

|[1,bc,a(b+c)],[1,ca,b(c+a)],[1,ab,c(a+b)]|=0

Delta=det[[1,a,bc1,b,ca1,c,ab]]