`|[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)

Prove that: 1/(bc+ca+ab)|[a, b, c],[a^2, b^2, c^2], [bc, ca, ab]|=(b-c),(c-a),(a-b)

Prove that |(1,1,1),(bc,ca,ab),(b+c, c+a, a+b)| = (a-b)(b-c)(c-a)

Match the following from List - I to List - II {:("List-I","List-II"),((I)|{:(1,1,1),(a,b,c),(bc,ca,ab):}|=,(a)(a-b)(b-c)(c-a)),((II)|{:(a,b,c),(a^(2),b^(2),c^(2)),(a^(3),b^(3),c^(3)):}|=,(b)(a-b)(b-c)(c-a)abc),((III)|{:(1,1,1),(a,b,c),(a^(3),b^(3),c^(3)):}|=,(c)(a-b)(b-c)(c-a)(a+b+c)):}

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

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

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

Prove that: {:|(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)