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

det[[a,a^(2),bcb,b^(2),cac,c^(2),ab]]=(a-b)(b-c)(c-a)(ab+bc+ca)

Prove the following : |{:(a,b,c),(a^(2),b^(2),c^(2)),(bc,ca,ab):}|=|{:(a,a^(2),bc),(b,b^(2),ca),(c,c^(2),ab):}|=(ab+bc+ca)(a-b)(b-c)(c-a) .

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

prove that , |{:(a,a^2,a^3+bc),(b,b^2,b^3+ca),(c,c^2,c^3+ab):}|=(a-b)(b-c)(c-a)(abc+bc+ca+ab)

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

Prove that |(1,a^2,bc),(a,b^2,ca),(1,c^2,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)

Prove that abs((a,b,c),(a^2,b^2,c^2),(bc,ca,ab))=(a-b)(b-c)(c-a)(ab+bc+ca)

Prove that: 1/(bc+ca+ab)|[a, b, c],[a^2, b^2, c^2], [bc, ca, ab]|=(b-c),(c-a),(a-b)

Without expanding, prove the following |(a,b,c),(a^2,b^2,c^2),(bc,ca,ab)|=(a-b)(b-c)(c-a)(ab+bc+ca)