How to prove that $\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}=4$? Using the Cardano formula, one can show that $\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}$ is a real root of the depressed cubic
$f(x)=x^3-6x-40$. Actually, one can show by the calculating the determinant that this is the only real root. On the other hand, by the rational root theorem, one can see that the possible rational root must be a factor of 40 and one can check that $f(4)=0$. Therefore, by uniqueness of the real root, one must have
$$
\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}=4\tag{1}
$$
I had the observation above when I solved the cubic equation $x^3-6x-40=0$. My question is as follows:

without referring to the unique real root of the cubic, can we show (1) directly?


[An attempt.]
When taking the cube on both sides of (1) and simplifying further, I ended up with (1) again. 
 A: Well, $(2\pm\sqrt2)^3=20\pm14\sqrt2$, that is $\sqrt[3]{20\pm14\sqrt2}
=2\pm\sqrt2$. Therefore
$$\sqrt[3]{2+14\sqrt2}+\sqrt[3]{20-14\sqrt2}=2+\sqrt2+2-\sqrt2=4.$$
A: Can we express $\sqrt[3]{20+14\sqrt2}$ as $a+b\sqrt2$, with $a,b\in\mathbb Z$? In other words, are thre integers $a$ and $b$ such that $(a+b\sqrt2)^3=20+14\sqrt2$? Note that\begin{align}(a+b\sqrt2)^3=20+14\sqrt2&\iff a^3+3\sqrt2a^2b+6ab^2+2\sqrt2b^3=20+14\sqrt2\\&\iff a^3+6ab^2+(3a^2b+2b^3)\sqrt2=20+14\sqrt2.\end{align}Therefore, it is enough to have$$a^3+6ab^2=20(\iff a(a^2+6b^2)=20)\text{ and }3a^2b+2b^3=14(\iff b(3a^2+2b^2)=14).$$It is easy to see that you can take $a=2$ and $b=1$. Therefore, $\sqrt[3]{20+14\sqrt2}=2+\sqrt2$ and it is now easy to see that $\sqrt[3]{20-14\sqrt2}=2-\sqrt2$. Therefore$$\sqrt[3]{20+14\sqrt2}+\sqrt[3]{20-14\sqrt2}=2+\sqrt2+2-\sqrt2=4.$$
A: If you don't spot the values of these cube roots and don't feel like computing them, you can try this. Define $a_\pm:=(20\pm14\sqrt{2})^{1/3},\,s:=\sum_\pm a_\pm$ so $\sum_\pm a_\pm^3=40$ and $\prod_\pm a_\pm=8^{1/3}=2$, so $s^2=\frac{40}{s}+3a_+a_-=\frac{40}{s}+6$. Hence $0=s^3-6s-40=(s-4)(s^2+4s+10)$ has only one real root, $4$.
