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

|[1,a, bc] ,[1, b, ca], [1, c, 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)

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

Prove that |[1,1,1] , [a,b,c] ,[a^2-bc, b^2-ca, c^2-ab]|=0

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

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

|[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]]

Show without expanding at any stage that: | [1,bc, a(b+c)],[1,ca, b(c+a)],[1,ab, c(a+b)]|=0

" if " |{:(1,,1,,1),(a,,b,,c),(bc,,ca,,ab):}|=|{:(1,,1,,1),(a,,b,,c),(a^(3),,b^(3),,c^(3)):}| where a,b,c are distinct positive reals then the possible values of abc is // are