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

Show that abs [[1,a,bc],[1,b,ac],[1,c,ab]]=(a-b)(b-c)(c-a)

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

Prove that: {:|(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)

Using properties of determinants , prove that |{:(1,a,bc),(1,b,ca),(1,c,ab):}|=(a-b)(b-c)(c-a)

Using properties of determinants prove the following. abs[[1,a,bc],[1,b,ca],[1,c,ab]]=(a-b)(b-c)(c-a)

Prove that |(1,a^2,bc),(a,b^2,ca),(1,c^2,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)