<s>[INST] Evaluation of $\sqrt[3]{40+11\sqrt{13}}+\sqrt[3]{40-11\sqrt{13}}$ 
Evaluate
$$\sqrt[3]{40+11\sqrt{13}}+\sqrt[3]{40-11\sqrt{13}}$$

The solution is $5$.
Suppose $\sqrt[3]{40+11\sqrt{13}}=A, \sqrt[3]{40-11\sqrt{13}}=B$
We have
$$A^3+B^3=80, A^3-B^3=22\sqrt{13}$$
Two unknowns, two equations, so we should be able to solve for $A+B$ (what we want).
How can I do that?
 [/INST] Let $x = \sqrt[3]{40 + 11\sqrt{13}} + \sqrt[3]{40 - 11\sqrt{13}}$. Then we have that
\begin{align*}
x^{3} = 80 + 9x & \Longleftrightarrow x^{3} - 9x - 80 = 0\\\\
& \Longleftrightarrow (x^{3} - 25x) + (16x - 80) = 0\\\\
& \Longleftrightarrow x(x^{2} - 25) + 16(x  - 5) = 0\\\\
& \Longleftrightarrow x(x + 5)(x - 5) + 16(x - 5) = 0\\\\
& \Longleftrightarrow (x^{2} + 5x + 16)(x - 5) = 0\\\\
& \Longleftrightarrow x = 5
\end{align*}
and we are done.
Hopefully this helps !
</s>