`|[1,a,a^2-bc],[1,b,b^2-ca],[1,c,c^2-ab]|`=0

Evaluate the following: |[1,,a^2-bc],[1, b,b^2-ac],[1,c,c^2-ab]|

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 that |[a^2+1,ab,ac],[ab,b^2+1,bc],[ac,bc,c^2+1]|=1+a^2+b^2+c^2

The value of |(a,a^(2) - bc,1),(b,b^(2) - ca,1),(c,c^(2) - ab,1)| , is

Using properties of determinant show that: |[1 , a , bc] , [1 , b , ca] , [1 , c , a b]|=(a-b)(b-c)(c-a)

Without expanding the determinant, prove that |(a,a^2,bc),(b,b^2,ca),(c,c^2,ab)|=|(1,a^2,a^3),(1,b^2,b^3),(1,c^2,c^3)|

Without expanding determinant at any stage evaluate |(1,1,1),(a,b,c),(a^(2)-bc,b^(2)-ca,c^(2)-ab)|

Without expanding at any stage, prove that |{:(1,a,a^(2)),(1,b,b^(2)),(1,c,c^(2)):}|=|{:(1,a,bc),(1,b,ca),(1,c,ab):}|

Using properties of determinants, prove the following: |[a^2 + 1,ab, ac], [ab,b^2 + 1,b c],[ca, cb, c^2+1]|=1+a^2+b^2+c^2

abs([-a^2,ab,ac],[ba,-b^2,bc],[ca,cb,-c^2]) = 4a^2.b^2.c^2