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

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

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

Without expanding, prove that : |{:(1, bc, a(b+ c) ),(1, ca, b ( c+ a) ),(1, ab , c( a+ b)):}|=0 .

|[1,bc,a(b+c)],[1,ca,b(c+a)],[1,ab,c(a+b)]|=0

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

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

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

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

Prove the following : [[1,bc,a(b+c)],[1,ca,b(c+a)],[1,ab,c(a+b)]] =0