Q. If a+ b+ c=0, where a≠b≠c, then a22a2+bc+b22b2+ac+c22c2+ab is equal to
(A) 0 , (B) 1, (C) -1 , (D) abc
1a2a2+(a)(a)+bc + b2b2+(b)(b)+ca + c2c2+(c)(c)+ab
= a2a2-a(b+c)+bc + b2b2-b(c+a)+ca + c2c2-c(a+b)+ab
= a2(a-b)(a-c) + b2(b-c)(b-a) + c2(c-a)(c-b)
= - a2(b-c)(a-b)(b-c)(c-a) - b2(c-a)(a-b)(b-c)(c-a) - c2(a-b)(a-b)(b-c)(c-a)
= - a2b - a2c + b2c - b2a + c2a - c2b(a-b)(b-c)(c-a)
= - a2b + a2c - b2c + b2a - c2a + c2b- a2b + a2c - b2c + b2a - c2a + c2b
= 1
1the second last step of bipin sir's can be simplified
by putting a=b
the expresion reduces to 0
hence a-b is a factor
since expression is symmetric and cyclic the same holds true for b-c and c-a
hence answer is 1[1]
39xyz...the condition is given a≠b≠c...kaise assume karen phir?
