|[a-b,b-c,c-a],[b-c,c-a,a-b],[c-a,a-b,b-c]|=0

Using the property of determinants and without expanding {:[( a-b,b-c, c-a),( b-c,c-a,a-b),( c-a,a-b,b-c)]:} =0

Using the property of determinants and without expanding {:[( a-b,b-c, c-a),( b-c,c-a,a-b),( c-a,a-b,b-c)]:} =0

Prove that: |[a+b, b+c, c+a],[b+c,c+a,a+b],[c+a,a+b,b+c]|=2|[a,b,c],[b,c,a],[c,a,b]|

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

Prove that : |[a+b+c,-c,-b],[-c, a+b+c, -a],[-b,-a,a+b+c]|= 2(a+b)(b+c)(c+a)

Show that |[b-c,c-a, a-b],[ c-a, a-b,b-c],[ a-b,b-c,c-a]| = 0 .

Show that abs{:(b-c,c-a,a-b),(c-a,a-b,b-c),(a-b,b-c,c-a):}=0.

If |(a+b,b+c,c+a),(b+c,c+a,a+b),(c+a,a+b,b+c)|=k|(a,b,c),(b,c,a),(c,a,b)| then k=

|(a,b-c,c+b),(a+c,b,c-a),(a-b,b+a,c)|=