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

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

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)):}

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]| = (a-b)(b-c)(c-a)

Prove that: {:|(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)

Prove that {:[(1,ab,a+b),(1,bc,b+c),(1,ca,c+a):}]=(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)