`|[a,b,c],[a-b,b-c,c-a],[b+c,c+a,a+b]|=a^3+b^3+c^3-3abc`

Prove the identities: |[a, b, c],[ a-b,b-c,c-a],[ b+c,c+a, a+b]|=a^3+b^3+c^3-3a b c

det[[a,b,ca-b,b-c,c-ab+c,c+a,a+b]]=a^(3)+b^(3)+c^(3)-3abc

Show that abs((a,b,c),(a-b,b-c,c-a),(b+c,c+a,a+b)) = a^3+b^3+c^3 -3abc.

Show that |[b+c,c+a,a+b] , [a+b,b+c,c+a] , [a,b,c]|=a^3+b^3+c^3-3abc

Prove: \ |(a, b, c),( a-b,b-c,c-a),( b+c,c+a, a+b)|=a^3+b^3+c^3-3a b c

|(1,b,c),(a-b,b-c,c-a),(b+c,c+a,a+b)|=a^(3)+b^(3)+c^(3)-3ab

Prove the identities: det[[a-b,b-c,c-ab+c,c+a,a+b]]=a^(3)+b^(3)+c^(3)-3abcdet[[a-b,b-c,c-ab+c,c+a,a+b]]=a^(3)+b^(3)+c^(3)-3abc

if abc!=0 and if |[a,b,c],[b,c,a],[c,a,b]|=0 then (a^3+b^3+c^3)/(abc)=

Prove the following: [[b+c,a+b,a],[c+a,b+c,b],[a+b,c+a,c]] = a^3+b^3+c^3-3abc

Consider the matrix A=[[a,b,c],[b,c,a],[c,a,b]] show that abs[A]=-(a^3+b^3+c^3-3abc)