Find $S$ where $S=\sqrt[3] {5+2 \sqrt {13}}+\sqrt[3]{5-2 \sqrt {13}}$, why am I getting an imaginary number? $\large S=\sqrt[3] {5+2 \sqrt {13}}+\sqrt[3]{5-2 \sqrt {13}}$ 
Multiplying by conjugate:
$\large S=\dfrac {-3}{\sqrt[3] {5+2 \sqrt {13}}-\sqrt[3]{5-2 \sqrt {13}}}$
From the original:
$\large S-2\sqrt[3]{5-2 \sqrt {13}}
=\sqrt[3] {5+2 \sqrt {13}}-\sqrt[3]{5-2 \sqrt {13}}$
Substituting:
$\large S=\dfrac{-3}{S-2\sqrt[3]{5-2 \sqrt {13}}}$
This leads to a quadratic equation in $\large S$ which I checked in wolframalpha and I got imaginary solutions. Why does this happen? I am not looking for an answer telling me how to solve this problem, I just want to know why this is wrong. Thanks.

For simplicity, we'll let $S = a+b$ where $a=\sqrt[3]{5+2\sqrt{13}}$ and $b=\sqrt[3]{5-2\sqrt{13}}$.
Note that $S^3=a^3+3a^2b+3ab^2+b^3=5+2\sqrt{13}+5-2\sqrt{13}+3a^2b+3ab^2$.
$S^3+9S-10=3a^2b+3ab^2+9a+9b=3a(ab+3)+3b(ab+3)=3(a+b)(ab+3)=3S(ab+3)$. 
Now, $ab=\sqrt[3]{(5+2\sqrt{13})(5-2\sqrt{13})}=\sqrt[3]{25-52}=\sqrt[3]{-27}=-3$. So $ab+3=0$. Thus, we see that $S$ is a root of the cubic equation $x^3+9x-10$. It is clear that $1$ is a root so the cubic is equal to $(x-1)(x^2+x+10)$. By the Quadratic Formula, the other two roots are imaginary. Thus $S=1$.
