The sum $\sqrt[3] {5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}$ equals 
$\text{(A)} \ \frac 32 \qquad  \text{(B)} \ \frac{\sqrt[3]{65}}{4} \qquad  \text{(C)} \ \frac{1+\sqrt[6]{13}}{2} \qquad  \text{(D)}\ \sqrt[3]{2}\qquad \text{(E)}\ \text{none of these}$
Lets set our original expression equal to $x$. So $\sqrt[3] {5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}} = x$. Cubing this gives us $x^3 = \left(\sqrt[3] {5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\right)^3 = 5 + 2\sqrt{13} + 5 - 2\sqrt{13} + 3\left(\sqrt[3] {5+2\sqrt{13}}*\sqrt[3]{5-2\sqrt{13}}\right)\left(\sqrt[3] {5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\right) = 10 - 9x$
So we have $x^3 + 9x - 10 = 0$. We can easily see that 1 is a root of this polynomial. By synthetic division, the new polynomial is $x^2 + x + 10$, which has no real roots. Thus $\sqrt[3] {5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}} = 1$. Since 1 is not A-D, our answer is $\fbox{E}$.