It is given that
\(\begin{aligned}
& \frac{3}{(x-1)\left(x^2+x+1\right)}=\frac{1}{x-1}-\frac{x+2}{x^2+x+1} \\
& =f_1(x)-f_2(x) \\
& \text { and } \frac{x+1}{(x-1)^2\left(x^2+x+1\right)}=A f_1(x)+\left(B+\frac{D}{x-1}\right) \\
& f_2(x)+\frac{C}{(x-1)^2}
\end{aligned}\)
From above informations, it is necessary, that
\(\begin{aligned}
f_1(x) & =\frac{1}{x-1} \\
\text { and } \quad f_2(x) & =\frac{x+2}{x^2+x+1}
\end{aligned}\)
So,
\(\begin{array}{r}
\frac{x+1}{(x-1)^2\left(x^2+x+1\right)}=\frac{A}{x-1}+\frac{(B x-B+D)(x+2)}{\left(x^2+x+1\right)(x-1)} \\
+\frac{C}{(x-1)^2}
\end{array}\)
\(\begin{aligned}
\Rightarrow x+1 & =A(x-1)\left(x^2+x+1\right)+B(x-1) \\
& (x+2)(x-1)+D(x+2)(x-1)+C\left(x^2+x+1\right)
\end{aligned}\)
On comparing the coefficient, we get coefficient of \(x^3=0 \Rightarrow A+B=0\) coefficient of \(x^2=0 \Rightarrow A-A+2 B-2 B+D+C=0\)
\(\begin{array}{llll}
\Rightarrow & D+C =0 \\
\therefore & A+B+C+D & =0
\end{array}\)
Hence, option (3) is correct.