If `|[a,a^(2),1+a^(3)],[b,b^(2),1+b^(3)],[c,c^(2),1+c^(3)]|=0" and "|[a,a^(2),1],[b,b^(2),1],[c,c^(2),1]|!=0` then show that `abc =-1`

If [[a,a^(2),a^(3)-1b,b^(2),b^(3)-1c,c^(2),c^(3)-1]]=0

|[a,a^2,a^3-1],[b,b^2,b^3-1],[c,c^2,c^3-1]|=0 prove that abc= I

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 , a ^(2), 1+ a ^(3)), ( b , b^(2), 1+ b ^(3)), ( c ,c ^(2), 1 + c ^(3)):}|=0 and vectors (1, a,a ^(2)), (1, b, b ^(2)) and (1, c, c^(2)) are non-coplanar, then the value of abc +1 is

|(1,a^(2)+bc,a^(3)),(1,b^(2)+ac,b^(3)),(1,c^(2)+ab,c^(3))|=-(a-b)(b-c)(c-a)(a^(2)+b^(2)+c^(2))

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

If |{:(a,,a^(2),,1+a^(3)),(b,,b^(2),,1+b^(3)),(c,,c^(2),,1+c^(3)):}|=0 and the vectors overset(to)(A) =(1, a, a^(2)) , overset(to)(B) = (1, b, b^(2)) , overset(to)(C )(1,c,c^(2)) are non-coplanar then the product abc = ….