Is $\left(45+29\sqrt{2}\right)^{1/3} + \left(45-29\sqrt{2}\right)^{1/3}$ an integer? The problem is the following:

Prove that this number
  $$x = \left(45+29\sqrt{2}\right)^{1/3} + \left(45-29\sqrt{2}\right)^{1/3}$$
  is an integer. Show which integer it is.

I  thought that it has some relations with something like complex numbers such as the set $N(\sqrt{2})$ or the some kind of integer polynomial which has root $x$ and therefore show that it has integer solutions.

Given the hint that $(45+29\sqrt{2})^{1/3} + (45-29\sqrt{2})^{1/3}$ is an integer.
One should first investigate  whether
$$(45 + 29\sqrt{2})^{1/3} = a + b\sqrt{2}$$
for some integers $a, b$.
Notice $$(a+b\sqrt{2})^3 = a ( a^2 + 6 b^2 ) + (3a^2 + 2b^2)b\sqrt{2}$$
This suggest us to look for integer solutions for following equation:
$$\begin{cases} 
a( a^2 + 6b^2) &= 45\\
b(3a^2 + 2b^2) &= 29
\end{cases}
$$
Since $b | 29$ and $29$ is a prime, the most natural guess is $b = 1$. Substitute $b = 1$ in the second equation, we find
$a = \sqrt{\frac{29 - 2}{3}} = 3$ which also solves the first equation. This leads to
$$(45+29\sqrt{2})^{1/3} = 3 + \sqrt{2} \implies
(45-29\sqrt{2})^{1/3} = 3 - \sqrt{2}\\ \implies
(45+29\sqrt{2})^{1/3} + (45-29\sqrt{2})^{1/3} = 6$$
