`Delta = |(a, a^2, 0),(1, 2a+b,(a+b)),(0, 1, 2a+3b)|` is divisible by `a+b` b. `a+2b` c. `2a+3b` d. `a^2`

Delta = |(a, a^2, 0),(1, 2a+b,(a+b)),(0, 1, 2a+3b)| is divisible by a. a+b b. a+2b c. 2a+3b d. a^2

Delta = |(a, a^2, 0),(1, 2a+b,(a+b)),(0, 1, 2a+3b)| is divisible by a. a+b b. a+2b c. 2a+3b d. a^2

|(a, a^2, 0), (1, 2a+b ,(a+b)), (0, 1, 2a+3b)| is divisible by a. a+b b. a+2b c. 2a+3b d. a^2

Let Delta=|(a,a^(2),0),(1,2a+b,(a+b)^(2)),(0,1,2a+3b)| then

=|a a^2 0 1 2a+b(a+b)0 1 2a+3b| is divisible by a+b b. a+2b c. 2a+3b d. a^2

Let f(a,b)=|{:(a,a^2,0),(1,(2a+b),(a+b)^2),(0,1,(2a+3b)):}| , then

The determinant Delta=|(a^2(a+b),a b,a c),(a b,b^2(a+k),b c),(a c,b c,c^2(1+k))| is divisible by

Prove: |(0,b^2a, c^2a),( a^2b,0,c^2b),( a^2c, b^2c,0)|=2a^3b^3c^3

Prove: |(0,b^2a, c^2a),( a^2b,0,c^2b),( a^2c, b^2c,0)|=2a^3b^3c^3

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