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.

It's probably better to cube $S$:
$$S^3 = (5 + 2\sqrt{13}) + 3\sqrt[3] {5+2 \sqrt {13}}\sqrt[3] {5- 2 \sqrt {13}}\bigg(\sqrt[3] {5+2 \sqrt {13}} + \sqrt[3] {5-2 \sqrt {13}}\bigg) + (5 - 2\sqrt{13})$$ 
$$= 10 + 3\sqrt[3] {5^2-(2 \sqrt {13})^2}S$$
$$= 10 -9S$$
So $S$ satisfies the equation $S^3 + 9S -10 = 0$. The polynomial $S^3 + 9S -10$ factorizes as $(S-1)(S^2 + S - 10) = 0$, so you expect that $S = 1$. 
But you still have to make sure $S$ is not one of the roots of $S^2 + S - 10 = 0$, given by ${\displaystyle -{1 \over 2} \pm {\sqrt{41} \over 2}}$. For this, note that the absolute value of each of these roots is greater than ${\sqrt{41} \over 2} - {1 \over 2} >  {\sqrt{36} \over 2} - {1 \over 2} > 2$. On the other hand $\sqrt[3] {5+2 \sqrt {13}}$ is less than $\sqrt[3] {5+2 \sqrt {16}} = \sqrt[3]{13} < 2$. The other term $\sqrt[3] {5-2 \sqrt {13}}$ is negative and of smaller absolute value, so we conclude that $0 < S < 2$, and the only possibility is that $S = 1$.
