<s>[INST] Proving that the number $\sqrt[3]{7 + \sqrt{50}} + \sqrt[3]{7 - 5\sqrt{2}}$ is rational I've been struggling to show that the number $\sqrt[3]{7 + \sqrt{50}} + \sqrt[3]{7 - 5\sqrt{2}}$ is rational. I would like to restructure it to prove it, but I can't find anything besides $\sqrt{50} =5 \sqrt{2}$. Could anybody give me some hints? Thanks in advance!
 [/INST] First, let us cube the number at hand:
$$\begin{align*}
x^3
&=7+\sqrt{50}+7-5\sqrt{2}+3(7+\sqrt{50})^\frac{2}{3}(7-5\sqrt{2})^\frac{1}{3}+3(7+\sqrt{50})^\frac{1}{3}(7-5\sqrt{2})^\frac{2}{3}\\
&=14+3((49-50)(7+\sqrt{50}))^\frac{1}{3}+3((49-50)(7-5\sqrt{2}))^\frac{1}{3}\\
&=14-3x
\end{align*}$$
Then, $x^3=14-3x \implies x^3+3x-14=0$.
The only real solution of this equation is $x=2$, which is a rational number.
QED.
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