<s>[INST] Is $\sqrt[3]{7+5 \sqrt{2}}+\sqrt[3]{20-14 \sqrt{2}}$ a rational number? Is there a way to show that
$$\alpha=\sqrt[3]{7+5 \sqrt{2}}+\sqrt[3]{20-14 \sqrt{2}}$$
is a rational number?
I found $\alpha=3$ from doing simplifications. But, I would like to known a different approach.
 [/INST] Note $\sqrt[3]{20-14 \sqrt{2}}=\sqrt2 \cdot \sqrt[3]{5 \sqrt{2}-7}$ and assume $\sqrt[3]{5\sqrt{2}\pm7}=p\sqrt2\pm q$. Then
$$\alpha
=(p\sqrt2+q)+\sqrt2(p\sqrt2-q)=(q+2p)+(p-q)\sqrt2
$$
For $a$ to be rational requires $p=q=\frac{\sqrt[3]{5 \sqrt{2}+7}}{\sqrt2+1}
= \sqrt[3]{\frac{ 5 \sqrt{2}+7}{(\sqrt2+1)^3 }}=1$ to be rational, which is indeed.
</s>