LetLHS=ytheny=(1x(a−b)(a−c))(1x(b−c)(b−a))(1x(c−a)(c−b))theny(a−b)(b−c)(c−a)=x−(b−c)x−(c−a)x−(a−b)=x−b+c−c+a−a+b=x0=1∴y=1=RHS(proved)
Question:- (1/x a-b)1/(a-c) . (1/x b-c)1/(b-a) . (1/x c-a) 1/(c-b) = ?
A) 0
B) 1
C) (a+b+c)
D) (a-b+c) 2