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

Without expanding, show that the value of each of the determinants is zero: |[1,a, 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

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

Compute the following: [[a^2+b^2, b^2+c^2],[a^2+c^2, a^2+b^2]] + [[2ab,2bc],[-2ac,-2ab]]

Using properties of determinants, prove the following |(a^2,ab,ac),(ab,b^2+1,bc),(ca,cb,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 determinants, prove the following |(a^2+1,ab,ac),(ab,b^2+1,bc),(ca,cb,c^2+1)|=1+a^2+b^2+c^2 .

Identify like terms in the following: abc, ab^(2)c, acb^(2), c^(2)ab, b^(2)ac, a^(2)bc, cab^(2)

If a,b,c are all distinct and |[a,a^3,a^4-1],[b,b^3,b^4-1],[c,c^3,c^4-1]| =0, show that abc(ab+bc+ac) = a+b+c

If a,b,c are all distinct and |[a,a^3,a^4-1],[b,b^3,b^4-1],[c,c^3,c^4-1]| =0, show that abc(ab+bc+ac) = a+b+c