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

Evaluate the following determinants: [[a-b,b-c,c-a],[b-c,c-a,a-b],[c-a,a-b,b-c]]

Evaluate abs([a-b,b-c,c-a],[b-c,c-a,a-b],[c-a,a-b,b-c])

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

Write the value of the following determinant abs{:(a-b, b-c, c-a),(b-c, c-a, a-b),(c-a, a-b, b-c):} .

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)

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

Show without expanding at any stage 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]|

Show that abs[[a+b,b+c,c+a],[b+c,c+a,a+b],[c+a,a+b,b+c]]=2abs[[a,b,c],[b,c,a],[c,a,b]]

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