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

If a^2+b^2+c^2+ab+bc+ca<=0 AA a, b, c in R then find the value of the determinant |[(a+b+2)^2, a^2+b^2, 1] , [1, (b+c+2)^2, b^2+c^2] , [c^2+a^2, 1, (c+a+2)^2]| : (A) abc(a^2 + b^2 +c^2) (B) 0 (C) a^3+b^3+c^3 + 3abc (D) 65

Value of the D=|(1/a,bc,a^(3)),(1/b,ca,b^(3)),(1/c,ab,c^(3))| is

Prove that det[[1,a^(2)+bc,a^(3)1,b^(2)+ca,b^(3)1,c^(2)+ca,c^(3)]]=-(a-b)(b-c)(c-a)(a^(2)+b^(2)+c^(2))det[[1,b^(2)+ca,b^(3)1,c^(2)+ca,c^(3)]]=-(a-b)(b-c)(c-a)(a^(2)+b^(2)+c^(2))

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

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)):}|

If a^(2) + b^(2) + c^(3) + ab + bc + ca le 0 for all, a, b, c in R , then the value of the determinant |((a + b +2)^(2),a^(2) + b^(2),1),(1,(b +c + 2)^(2),b^(2) + c^(2)),(c^(2) + a^(2),1,(c +a +2)^(2))| , is equal to

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

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