Answer
$\frac{2x+4}{x^3-1}=\frac{2}{x-1}-\frac{2x+2}{x^2+x+1}$
Work Step by Step
1. Factor $x^3-1=(x-1)(x^2+x+1)$
2. Assume $\frac{2x+4}{x^3-1}=\frac{A}{x-1}+\frac{Bx+C}{x^2+x+1}$
3. Combine the right side of above to get:
$\frac{A}{x-1}+\frac{Bx+C}{x^2+x+1} =\frac{Ax^2+Ax+A+Bx^2-Bx+Cx-C}{(x-1)(x^2+x+1)} =\frac{(A+B)x^2+(A-B+C)x+(A-C)}{(x-1)(x^2+x+1)}$
4. Compare with the left side of step2, we have
$\begin{cases} A+B=0 \\ A-B+C=2 \\ A-C=4 \end{cases}$
5. Add the 2nd and 3rd equations to get $2A-B=6$
6. Add the above to the 1st equation to get $3A=6$, thus $A=2$
7. We can find $B=-2$ and $C=A-4=-2$
8. Thus $\frac{2x+4}{x^3-1}=\frac{2}{x-1}-\frac{2x+2}{x^2+x+1}$
