Let \( f(x)=a x^{2}+b x+c \). If \( x_{1}, x_{2}, x_{3} \) are three distinct real numbers such that \( f\left(x_{1}\right)=f\left(x_{2}\right)=f\left(x_{3}\right)=0 \), then \( f(x) \equiv 0 \), i.e. \( f(x)=0 \forall x \in \mathbf{R} \).
If \( g(x)=\frac{x_{2} x_{3}\left(x-x_{1}\right)^{2}}{\left(x_{1}-x_{2}\right)\left(x_{1}-x_{3}\right)}+\frac{x_{3} x_{1}\left(x-x_{2}\right)^{2}}{\left(x_{2}-x_{3}\right)\left(x_{2}-x_{1}\right)} \)
\[
+\frac{x_{1} x_{2}\left(x-x_{3}\right)^{2}}{\left(x_{3}-x_{1}\right)\left(x_{3}-x_{2}\right)}
\]
then \( g(x) \) is identically equal to
(a) 1
(b) \( x \)
(c) \( x^{2} \)
(d) \( x^{2}-x+1 \)
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