<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] All of the other approaches here are based on guessing that the expressions under the cube root signs are perfect cubes.
Here's an approach that doesn't assume that.
Let $$a = \sqrt[3]{7 + \sqrt{50}} = \sqrt[3]{7 + 5\sqrt{2}}, \quad b = \sqrt[3]{7 - 5\sqrt{2}}.$$
We are trying to evaluate $s = a + b$. We have
$$a^3 = 7 - 5\sqrt{2}, \quad b^3 = 7 + 5\sqrt{2}.$$
So
$$a^3 + b^3 = 14, \quad a^3 b^3 = -1.$$
Since $a$ and $b$ are real, the second equation implies $ab = -1$. Now we find
$$14 = a^3 + b^3 = (a+b)^3 - 3ab(a + b) = s^3 +3s.$$
This cubic equation has only one real solution, $s = 2$, which can be found using the rational root theorem.
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