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Suppose a,b,c are three distinct real numbers.Let P(x)= (x-b)(x-c) / (a-b)(a-c) + (x-c)(x-a) / (b-c)(b-a) + (x-a)(x-b) / (c-a)(c-b) When simplified P(x) becomes
P(x)=(x-b)(x-c)/(a-b)(a-c) + (x-c)(x-a)/(b-c)(b-a) + (x-a)(x-b)/(c-a)(c-b)=(x-a)(x-b)(x-c)/(a-b)(b-c)(c-a)[(c-b)/(x-a) +(a-c)/(x-b)+(b-a)/(x-c)]Thanks & RegardsRinkoo GuptaAskIITians Faculty
let x=a,then, substituting in the given equation,p(x)=1+0+0=1similarly, p(x where x=b)=1, p(x where x=c)=1thus, generalized answer can be taken as 1.
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