The value of `|[0,a-b,a-c],[b-a,0,b-c],[c-a,c-b,0]|`is

Prove that : |{:(0,a-b,a-c),(b-a,0,b-c),(c-a,c-b,0):}|=0

Prove that: |[0,a,-b],[-a,0,-c],[ b, c,0]|=0 .

Find the number of real root of the equation |[0,x-a, x-b],[ x+a,0,x-c],[ x+b, x+c,0]|=0,a!=b!=c and b(a+c)> a c

Using the property of determinants and without expanding, prove that: |[a-b,b-c,c-a],[ b-c,c-a ,a-b ],[c-a ,a-b,b-c]|=0

If a,b,c are in G.P. then the value of |(a,b,a+b),(b,c,b+c),(a+b,b+c,0)|= (A) 1 (B) -1 (C) a+b+c (D) 0

If a ,b ,c are different, then the value of |[0,x^2-a, x^3-b],[ x^2+a,0,x^2+c],[ x^4+b, x-c,0]| is a. c b. a c. b d. 0

If a ,b ,c are different, then the value of |[0,x^2-a, x^3-b], [x^2+a,0,x^2+c], [x^4+b, x-c,0]| is a. b b. c c. b d. 0

If a >0 and discriminant of a x^2+2b x+c is negative, then |[a,b,ax+b],[b,c,bx+c],[ax+b,bx+c,0]| is a. +v e b. (a c-b)^2(a x^2+2b x+c) c. -v e d. 0

If a >0 and discriminant of a x^2+2b x+c is negative, then |[a,b,ax+b],[b,c,bx+c],[ax+b,bx+c,0]| is a. +v e b. (a c-b)^2(a x^2+2b x+c) c. -v e d. 0

Show that |[b-c,c-a, a-b],[ c-a, a-b,b-c],[ a-b,b-c,c-a]| = 0 .