" (16) Prove it कि- "|[a,b,c],[a^(2),b^(2),c^(2)],[bc,ca,ab]|=(a-b)(b-c)(c-a)(ab+bc+a)

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

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)

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

Prove that the following. [[a,b,c],[a^2,b^2,c^2],[bc,ca,ab]] =(b-c)(c-a)(a-b)(bc+ca+ab)

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

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