Factoring the numerator using the sum of cubes identity and denominator using the difference of squares identity gives $$\dfrac{(x+1)(x^{2}-x+1)}{(x+1)(x-1)}$$
Cancelling out a factor of $x+1$ from the numerator and denominator gives $$\dfrac{(x^{2}-x+1)}{(x-1)}$$
Plugging in $x= -1$ gives $\dfrac{3}{-2}$ or $\fbox{E}$.