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!

Solution 1:

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.

Solution 2:

This should be an easy way (I converted my comment into an answer) :

Since $$7±5\sqrt 2=1±3\sqrt2 +6±2\sqrt 2=(1±\sqrt 2)^3,$$ we have $$\sqrt[3]{7+5\sqrt 2}+\sqrt[3]{7-5\sqrt 2}=(1+\sqrt 2)+(1-\sqrt 2)=2.$$