Solution:

$\begin{array}{c}{x^{3}} \\ {x ^ { 3 } + {1}{)}\hspace{-0.35em}\overline{\hspace{0.5em} x^4}} \\ {x^{4}+x} \\ {-\quad-} \\ \hline-x\end{array}$

$\begin{array}{l}{\frac{x^{4}}{x^{3}+1}=x-\frac{x}{x^{3}+1}} \\ {\frac{x}{x^{3}+1}=\frac{x}{(x+1)\left(x^{2}-x+1\right)}=\frac{A}{x+1}+\frac{B x+C}{x^{2}-x+1}} \\ {\therefore x=\left(x^{2}-x+1\right) A+(x+1)(B x+C)} \\ {\text { Put } x=-1} \\ {\therefore-1=3 A} \\ {\therefore A=\frac{-1}{3}}\end{array}$

$\begin{array}{l}{\text { Put } x=0} \\ {0=(1) A+(1) C} \\ {0=\frac{-1}{3}+C} \\ {\therefore C=\frac{1}{3}}\end{array}$

$Put x=1$

$\therefore 1=(1) A+2(B+C)$

$\therefore 1=\frac{-1}{3}+2 B+\frac{2}{3}$

$\therefore 1-\frac{1}{3}=2 B$

$\therefore \frac{2}{3}=2 B$

$\therefore B=\frac{1}{3}$

$\therefore \frac{x}{(x+1)\left(x^{2}-x+1\right)}=\frac{\frac{-1}{3}}{x+1}+\frac{\frac{1}{3} x+\frac{1}{3}}{x^{2}-x+1}$

$\frac{x^{4}}{x^{3}+1}=x-\frac{\frac{-1}{3}}{x+1}+\frac{\frac{1}{3} x+\frac{1}{3}}{x^{2}-x+1}$