If the sum of two numbers is $1$ and their product is $1$, then the sum of their cubes is:
$\text{(A)} \ 2 \qquad \text{(B)} \ - 2 - \frac {3i\sqrt {3}}{4} \qquad \text{(C)} \ 0 \qquad \text{(D)} \ - \frac {3i\sqrt {3}}{4} \qquad \text{(E)} \ - 2$
Let the two numbers be $a,b$; then $a^3 + b^3 = (a+b)(a^2 - ab + b^2) = (a+b)[(a+b)^2 -3ab]$ $= (1)[1^2 - 3 \cdot 1] = -2 \Rightarrow \mathrm{(E)}$.