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.
 A: Hint: Find $a>0$ and $b>0$ such that
$$
(a+b\sqrt{2})^3=45+29\sqrt{2},\quad (a-b\sqrt{2})^3=45-29\sqrt{2}.\tag{$*$}
$$
Show that this implies $a^3+6ab^2=45$ and $a^2-2b^2=7$. From this, you can find $a$ and $b$ and your answer is $2a$. (Further hint: consider $(a^3+6ab^2)+3a(a^2-2b^2)=45+21a)$. )
One might argue that this approach is not rigorous but it actually is, because once you have $a$ and $b$, you can check that ($*$) works by plugging in directly.
A: Let $u=\sqrt[3]{45+29\sqrt2}$ and $v=\sqrt[3]{45-29\sqrt2}$; so, $x=u+v$. We have
$$
u^3 + v^3=90,\\
uv = \sqrt[3]{45+29\sqrt2}\cdot \sqrt[3]{45-29\sqrt2} = \sqrt[3]{45^2-29^2\cdot2} = \sqrt[3]{343} = 7
$$
But
$$
u^3 + v^3 = (u+v)(u^2 - uv + v^2) = (u+v)((u+v)^2 - 3uv),
$$
and
$$
90 = x(x^2 - 21).
$$
Since $x$ must be integer, check divisors of $90$. $x^2 - 21 > 0$, so $x \ge 5$. Try $x=6$:
$$
90 = 6(6^2-21) = 6\cdot 15.
$$
A: 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
$$(45+29\sqrt{2})^{1/3} = 3 + \sqrt{2} \implies
(45-29\sqrt{2})^{1/3} = 3 - \sqrt{2}\\ \implies
(45+29\sqrt{2})^{1/3} + (45-29\sqrt{2})^{1/3} = 6$$
A: Let $a,b$ be the first and second term of the sum then $x = a+b$. Observe that: $ab = 7 \to x^3 = (a+b)^3 = a^3+b^3 + 3ab(a+b)=90+3\times 7\times x \to x^3-21x-90 = 0\to (x-6)(x^2+6x+15)=0\to (x-6)((x+3)^2+6)=0\to x = 6.$
