|[3a,-a+b,-a+c],[-b+a,3b,-b+c],[-c+a,-c+b,3c]|=3(a+b+c)(ab+bc+ca)

Show that [[3a,-a+b,-a+c],[-b+a,3b,-b+c],[-c+a,-c+b,3c]] = 3(a+b+c)(ab+bc+ca)

Show that: ,3a,-a+b,-a+c-b+a,3b,-b+c-c+a,-c+b,3c]|=3(a+b+c)(ab+bc+ca)

3a,-a+b,-a+ca-b,3b,c-ba-b,3b,c-ba-c,b-c,3c]|=

Using properties of determinants, prove the following |(3a,-a+b,-a+c),(a-b,3b,c-a),(a-c,b-c,3c)|=3(a+b+c)(ab+bc+ca)

Prove that |(3a, -a+b, -a+c),(a-b, 3b, c-b),(a-c, b-c, 3c)| = 3(a+b+c)(ab+bc+ca) .

solve {:[( 3a,-a+b,-a+c),( -b+a, 3b,-b+c) ,( -c+a,-c+b,3c) ]:} = 3( a+b+c) ( ab+bc+ca)

Prove {:[( 3a,-a+b,-a+c),( -b+a, 3b,-b+c) ,( -c+a,-c+b,3c) ]:} = 3( a+b+c) ( ab+bc+ca)

|{:(" "3a," "-a+b,-a+c),(a-b," "3b," "c-b),(a-c," "b-c," "3c):}|=3(a+b+c)(ab+bc+ca)

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

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)