<s>[INST] 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.
 [/INST] Note the identity $$x^3+y^3+z^3=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)+3xyz$$
So that if $x+y+z=0$ we have $x^3+y^3+z^3=3xyz$
This is often useful with cubic expressions like this. Take $x=-S;\text{  } y=\sqrt[3] {5+2 \sqrt {13}};\text{  }z=\sqrt[3]{5-2 \sqrt {13}}$
Then we have$$-S^3+5+2\sqrt{13}+5-2\sqrt{13}=-3S\sqrt[3]{25-52}$$ whence $$S^3+9S-10=0$$
and you recover the cubic of which $S$ is a root. As it happens this is easy to factor, and you need to identify the root which corresponds to the expression you were given. I shall not repeat what Zarrax has put from this point on. But I think the trick is worth knowing, as it hugely simplifies the arithmetic.
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