Let a,b, and c be distinct real numbers. Simplify the expression (a−b)(a−c)(x+a)2+(b−a)(b−c)(x+b)2+(c−a)(c−b)(x+c)2.
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Let p(x)=(a−b)(a−c)(x+a)2+(b−a)(b−c)(x+b)2+(c−a)(c−b)(x+c)2.Then p(−a)=(a−b)(a−c)(−a+a)2+(b−a)(b−c)(−a+b)2+(c−a)(c−b)(−a+c)2=(b−a)(b−c)(b−a)2+(c−a)(c−b)(c−a)2=b−cb−a+c−bc−a=b−cb−a+b−ca−c=b−cb−c=1.Similarly, p(−b)=p(−c)=1. Since p(x)=1 for three distinct values of x, by the Identity Theorem, p(x)=1 for all x.