$-$ Segundo a identidade do enunciado, \begin{matrix} \Large{\frac{3}{x^3 + 1} = \frac{a}{x+1} + \frac{bx+c}{x^2 - x +1}} &,& \color{gray}{\large{\frac{(x^3 + 1) + 3}{x^3 + 1}}} &\color{gray}{=}& \color{gray}{1 + \Large{\frac{3}{x^3 + 1}}} \end{matrix} $\color{orangered}{Obs:}$ $\begin{matrix} x^3 + 1^3 &=& (x+1)(x^2 -x + 1) \end{matrix} $ \begin{matrix} \Large{\frac{3}{(x+1)(x^2 -x + 1)} = \frac{a(x^2 -x + 1) \ + \ (bx + c)(x+1)}{(x+1)(x^2 -x + 1)} } \end{matrix} Assim, \begin{matrix} x^2 (a+b) &+& x(b+c-a) &+&( a+c ) &=& 0x^2 &+& 0x &+& 3 \end{matrix} $-$ Igualando os coeficientes dos termos de mesma potência: \begin{matrix} \begin{cases} a+b &=& 0 \\ b+c-a &=& 0 \\ a+c &=& 3 \end{cases} &\Rightarrow& a= 1 &,& b = -1 &,& c = 2 \end{matrix} $-$ Portanto, \begin{matrix} a+b+c = 2 \\ \\ Letra \ (D) \end{matrix}