To be continuous at $x = - 1$, the value of $\frac {x^3 + 1}{x^2 - 1}$ is taken to be: 
$\textbf{(A)}\ - 2 \qquad \textbf{(B)}\ 0 \qquad \textbf{(C)}\ \frac {3}{2} \qquad \textbf{(D)}\ \infty \qquad \textbf{(E)}\ -\frac{3}{2}$
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 $\fbox{E}$.