<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] Let $(a+b\sqrt{2})^3=7+5\sqrt{2}$, then $(a-b\sqrt{2})^3=7-5\sqrt{2}$. This gives us
\begin{align}
a^3+6ab^2 & = 7\\
2b^3+3a^2b & = 5
\end{align} 
From the two equations listed above we get.
$$\frac{a(a^2+6b^2)}{b(2b^2+3a^2)}=\frac{7}{5}.$$
Now let $x=\frac{a}{b}$. Then
$$5x(x^2+6)=7(3x^2+2).$$
Finally we get
$$5x^3-21x^2+30x-14=0$$
Observe that $x=1$ is an obvious solution of this equation. So we can rewrite this as:
$$(x-1)(5x^2-16x+14)=0.$$
The quadratic factor has only non-real roots. Thus $x=1$ is the only real solution. This gives $a=b=1$. Thus the given expression is equal to $2$.
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