`[[a,a^2,(b+c)],[b,b^2, (a+c)],[c,c^2,(a+b)]] = (b-c)(c-a)(a-b)(a+b+c)`

Using properties of determinants, show that abs[[a,a^2,b+c],[b,b^2,c+a],[c,c^2,a+b]]=(b-c)(c-a)(a-b)(a+b+c)

Prove the identities: |[a^2,a^2-(b-c)^2,b c], [b^2,b^2-(c-a)^2,c a],[ c^2,c^2-(a-b)^2,a b]|=(a-b)(b-c)(c-a)(a+b+c)(a^2+b^2+c^2)

Prove the identities: |[a^2,a^2-(b-c)^2,b c], [b^2,b^2-(c-a)^2,c a],[ c^2,c^2-(a-b)^2,a b]|=(a-b)(b-c)(c-a)(a+b+c)(a^2+b^2+c^2)

Prove: |((b+c)^2, a^2, b c) ,((c+a)^2, b^2 ,c a),( (a+b)^2, c^2, a b)|=(a-b)(b-c)(c-a)(a+b+c)(a^2+b^2+c^2) .

Prove: |(a^2,a^2-(b-c)^2,b c), (b^2,b^2-(c-a)^2,c a),( c^2,c^2-(a-b)^2,a b)|=(a-b)(b-c)(c-a)(a+b+c)(a^2+b^2+c^2)

|{:(a,b,c),(a^2,b^2,c^2),(b+c,c+a,a+b):}|=(a-b)(b-c)(c-a)(a+b+c)

Prove that: {:|(a^2,a,b+c),(b^2,b,c+a),(c^2,c,ab)| = -(a+b+c)(a-b)(b-c)(c-a)

The determinant |[a^2, a^2-(b-c)^2,bc],[b^2,b^2-(c-a)^2,ca],[ c^2,c^2-(a-b)^2,ab]| is divisible by- a. a+b+c b. (a+b)(b+c)(c+a) c. a^2b^2c^2 d. (a-b)(b-c)(c-a)

Show that |(b+c,a,a^(2)),(c+a,b,b^(2)),(a+b,c,c^(2))|=(a+b+c)(a-b)(b-c)(c-a)