|[b+c,ab],[c+a,ca],[c1+b,bc]|=(a+b+c)(a-c)^(2)

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

Using the properties of determinants show that : |[[a^2, b^2, c^2],[bc,ca,ab],[a,b,c]]|=(a-b)(b-c)(c-a)(ab+bc+ca)

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

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

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

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 that |[(a+b)^(2),ca,bc],[ca,(b+c)^(2),ab],[bc,ab,(c+a)^(2)]|=2abc(a+b+c)^(3)

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

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