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

Evaluate: |[1,a,bc],[1,b,ca],[1,c,ab]|

Evaluate : triangle = |[1,a,bc],[1,b,ca],[1,c,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)):}

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

If A=|[1, 1, 1],[a, b, c],[ a^2,b^2,c^2]| , B=|[1,bc, a],[1,ca, b],[1,ab, c]| , then

Without expanding determinant show that |(xa,yb,zc),(a^2,b^2,c^2),(1,1,1)|=|(x,y,z),(a,b,c),(bc,ca,ab)|

Show that |[1+a,1,1],[1,1+b,1],[1,1,1+c]|=abc(1+1/a+1/b+1/c)=abc+bc+ca+ab

Show that |[1+a,1,1],[1,1+b,1],[1,1,1+c]| = abc(1+1/a+1/b+1/c) = abc+bc+ca+ab

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

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