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!

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.
