<s>[INST] Sum of cube roots of two conjugate quadratic integers makes an integer. Consider the following expression:
$$(20-\sqrt{392})^{1/3}+(20+\sqrt{392})^{1/3}.$$
This equals $4$, but how can I show this?
Note that I do not want to make use of the following line of reasoning: 4 is a solution to $x^3-6x-40=0$, that this cubic has exactly one solution, and that all solutions to the cubic
$x^3+px +q = 0$ are given by
$$(-q/2-((q/2)^2+(p/3)^3)^{0.5})^{1/3}+(-q/2+((q/2)^2+(p/3)^3)^{0.5})^{1/3}.$$
I have noted the method set out here, How can I show that this complicated expression with square and cube roots reduces to the value 7? , but implementing it does not work very well.
I attempted this and got more complicated expressions than I started off with! It doesn't result in a unique solution for $a$ and $b$.
I do think that this is the right approach, though, i.e. making use of the fact that if $\sqrt c$ is irrational and $a+b\sqrt c=d+e \sqrt f$ then $a=d$ and $b=e$. (How do you prove this fact, by the way?)
 [/INST] First note that $392=2^3\times7^2$ and so $\sqrt{392}=14\sqrt{2}$, and hence
$$\sqrt[3]{20+\sqrt{392}}=\sqrt[3]{20+14\sqrt{2}}.$$
Next, in the hope of finding a simple expression for the cube root, we compute
$$(20+14\sqrt{2})(20-14\sqrt{2})=20^2-2\times14^2=8,$$
which shows that
$$\sqrt[3]{20+14\sqrt{2}}\sqrt[3]{20-14\sqrt{2}}=\sqrt[3]{8}=2.$$
This suggests that perhaps there exist integers $a$ and $b$ such that
$$(a+b\sqrt{2})^3=20+14\sqrt{2},$$
and then certainly $a$ and $b$ should satisfy $a^2-2b^2=2$. An obvious choice is $a=2$ and $b=1$, and indeed
$$(2+\sqrt{2})^3=2^3+3\times2^2\times\sqrt{2}+3\times2\times\sqrt{2}^2+\sqrt{2}^3=20+14\sqrt{2},$$
and entirely analogously $(2-\sqrt{2})^3=20-14\sqrt{2}$. It follows that
$$\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}=(2+\sqrt{2})+(2-\sqrt{2})=4.$$

Alternatively, we could directly try to find integers $a$ and $b$ such that
$$(a+b\sqrt{2})^3=20+14\sqrt{2}.$$
This boils down to the two simultaneous cubic equations
\begin{eqnarray*}
20&=&a^3+6ab^2&=&a(a^2+6b^2),\\
14&=&3a^2b+2b^3&=&b(3a^2+2b^2),
\end{eqnarray*}
from which it quickly follows that $a$ and $b$ are positive divisors of $20$ and $14$, respectively, and it is not hard to check that $a=2$, $b=1$ is the only solution.
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