(x+b)(x+c)(b−a)(c−a)+(x+c)(x+a)(c−b)(a−b)+(x+a)(x+b)(a−c)(b−c)=1\frac{(x+b)(x+c)}{(b-a)(c-a)} + \frac{(x+c)(x+a)}{(c-b)(a-b)} + \frac{(x+a)(x+b)}{(a-c)(b-c)} = 1…

Problem

(x+b)(x+c)(b−a)(c−a)+(x+c)(x+a)(c−b)(a−b)+(x+a)(x+b)(a−c)(b−c)=1\frac{(x+b)(x+c)}{(b-a)(c-a)} + \frac{(x+c)(x+a)}{(c-b)(a-b)} + \frac{(x+a)(x+b)}{(a-c)(b-c)} = 1 is an identity.

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