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

Using properties of determinants, show the following: |[(b+c)^2,ab, ca],[ab,(a+c)^2,bc ],[ac ,bc,(a+b)^2]|=2abc(a+b+c)^3

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

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

|(a,b,c),(a^(2),b^(2),c^(2)),(bc,ca,ab)|=

Prove the identities: |{:(b^(2)+c^(2),,ab,,ac),(ab,,c^(2)+a^(2),,bc),(ca,,bc,,a^(2)+b^(2)):}|=4a^2b^2c^2

|[bc-a^2,ca-b^2,ab-c^2],[ca-b^2,ab-c^2,bc-a^2],[ab-c^2,bc-a^2,ca-b^2]|=|[a,b,c],[b,c,a],[c,a,b]|^2

|{:(a^2,ab,ac),(ab,b^2,bc),(ca,bc,c^2):}|= ?

If A=[(a^(2),ab,ac),(ab,b^(2),bc),(ac,bc,c^(2))] and a^(2)+b^(2)+c^(2)=1 , then A^(2)=