`|[1,a, a^2-b c], [1,b,b^2-a c],[1,c,c^2-a b]|=`

Show without expanding that |[1,a, a^2],[ 1,b,b^2],[ 1,c,c^2]|=|[1,b c, b+c],[1,c a, c+a],[1,a b, a+b]|

The value of the determinant |(1,a,a^2-bc),(1,b,b^2-ca),(1,c,c^2-ab)| is (A) (a+b+c),(a^2+b^2+c^2) (B) a^3+b^3+c^3-3abc (C) (a-b)(b-c)(c-a) (D) 0

Show that |[1,a,a^2],[1,b,b^2],[1,c,c^2]|=(a-b)(b-c)(c-a)

Prove: |(1,a, b c),(1,b ,c a),(1,c ,a b)|=|(1,a ,a^2),( 1,b,b^2),( 1,c,c^2)|

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

Value of |(1,a,a^2),(1,b,b^2),(1,c,c^2)| is (A) (a-b)(b-c)(c-a) (B) (a^2-b^2)(b^2-c^2)(c^2-a^2) (C) (a-b+c)(b-c+a)(c+a-b) (D) none of these

If |(a,a^2,1+a^3),(b,b^2,1+b^3),(c,c^2,1+c^2)|=0 and vectors (1,a,a^2),(1,b,b^2) and (1,c,c^2) are hon coplanar then the product abc equals (A) 2 (B) -1 (C) 1 (D) 0

If a!=b!=c such that |[a^3-1,b^3-1,c^3-1] , [a,b,c] , [a^2,b^2,c^2]|=0 then