1even im getting the same,
f(x) = x2 - (a+b+c)x + ab+bc+ac
\hspace{-16}$If $\bf{a,b,c\in\mathbb{R}}$ and $\bf{f(x)}$ is a Quadratic Polynomial such that\\\\ $\bf{\begin{Bmatrix} \bf{f(a)=bc} \\\\ \bf{f(b)=ca} \\\\ \bf{f(c)=ab} \end{Bmatrix}}$\\\\\\ Then $\bf{f(a+b+c)=}$
I am Getting f(a+b+c) = (ab+bc+ca)
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4 Answers
341P(x) = xf(x) -abc is a cubic with roots a,b,c and hence is equivalent to
x^3-x(a+b+c)+x^2(ab+bc+ca)-abc (bearing in mind that the constant term on both sides is -abc)
so that f(x) \equiv x^2-x(a+b+c)+(ab+bc+ca)
Its easily seen that f(a+b+c)=ab+bc+ca
