`|[1/a, a^2,b c],[1/b,b^2,c a],[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]|

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

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

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)|

If a b+b c+c a=0 , then what is the value of (1/(a^2-b c)+1/(b^2-c a)+1/(c^2-a b)) ? (a) 0 (b) 1 (c) 3 (d) a+b+c

If |[a, b, c], [a^(2), b^(2), c^(2)], [a^(3)+1, b^(3)+1, c^(2)+1]|=0 and the vectors given by A(1, a, a^(2)), B(1, b, b^(2)), C(1, c, c^(2)) are non-collinear, then abc=

If |(a,a^2,1+a^3),(b,b^2,1+b^3),(c,c^2,1+c^3)|=0 and the vectors A-=(1, a , a^2), B-=(1, b , b^2), C-=(1, c , c^2) are non-coplanar then the value of abc equal to

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