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

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

Using the property of determinants and without expanding,prove that: det[[1,bc,a(b+c)1,ca,b(c+a)1,ab,x(a+b)]]=0

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

Using the properties of determinants,prove det[[1,bc,bc(b+c)1,ca,ca(c+a)1,ab,ab(a+b)]]

Show that |{:(bc,b-c,1),(ca,c+a,1),(ab,a+b,1):}|=(a-b)(b-c)(c-a)

Show that |{:(bc,b+c,1),(ca,c+a,1),(ab,a+b,1):}|=(a-b)(b-c)(c-a)

Prove the following : [[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

Show without expanding at stage that : det[[1,ca,b(c+a)1,ab,c(a+b)]]=0

Delta=det[[1,a,bc1,b,ca1,c,ab]]