<s>[INST] Finding $x-\frac{1}{x}$, given $x^3 - \frac{1}{x^3} = 108+76\sqrt{2}$ 
If $x^3 - \dfrac{1}{x^3} = 108+76\sqrt{2}$, find the value of $x-\dfrac{1}{x}$.

Here's what I've tried so far.
$$\begin{align}
\left(x-\dfrac{1}{x}\right)^3&=x^3-\dfrac{1}{x^3}-3\left(x-\dfrac{1}{x}\right) \\
\rightarrow \quad \left(x-\dfrac{1}{x}\right)^3&=108+76\sqrt{2}-3\left(x-\dfrac{1}{x}\right) \\u:=x-\dfrac{1}{x} \quad\rightarrow \quad u^3+3u-108-76\sqrt{2}&=0
\end{align}$$
Got stuck here since I didn't know how to solve this cubic equation.
I also tried factorizing $x^3-\dfrac{1}{x^3}$.
$$\begin{align}x^3-\dfrac{1}{x^3}&=\left(x-\dfrac{1}{x}\right)\left(x^2+\dfrac{1}{x^2}+1\right) \\
&= \left(x-\dfrac{1}{x}\right)\left(x^2+\dfrac{1}{x^2}+2-1^2\right) \\
&= \left(x-\dfrac{1}{x}\right)\left(\left(x+\dfrac{1}{x}\right)^2-1^2\right) \\
&= \left(x-\dfrac{1}{x}\right)\left(x+\dfrac{1}{x}+1\right)\left(x+\dfrac{1}{x}-1\right)
\end{align}$$
Again, I didn't know what I could do with this.
 [/INST] Here is a suggestion to get an answer systematically - though not fully analytic.
Others have already noted that we need to solve for the desired $u = x- \frac{1}{x}$ in $$f(u) = u^3+3u- 108- 76\sqrt{2} = 0$$
Now observe that any expression of the form $y = a + b \sqrt{2}$, when taken to an integer power, reproduces this form, i.e. $y^n = a_n + b_n \sqrt{2}$. Since the constant part in $f(u)$ is also of this form,  we can write $u = a + b \sqrt{2}$, which gives
$$f(u) = a^3 +  6 a b^2 + 3 a -108 + \sqrt{2} (3 a^2 b +2  b^3 + 3  b - 76)   = 0 $$
This enables us to solve for the rational and the irrational parts separately, i.e.
$$a^3 +  6 a b^2 + 3 a -108  = 0\\
3 a^2 b +2  b^3 + 3  b - 76   = 0 $$
where $a$ and $b$ are rational numbers. Note that it is not clear beforehand whether $a$ and $b$ are integers, as other answers have tested some integer combinations.
Now this is again two coupled third-degree equations so one might ask what has been won (maybe one of you has a nice idea how to solve this analytically).
I introduce now a  way to manipulate these equations to come up with an iterative procedure with works without any knowledge or presupposition of the nature of  $a$ and $b$. Also, it does not need any roots (of quadratic or cubic equations) which have been used earlier but which can become cumbersome.
Writing equivalently
$$a^2 +  6  b^2 + 3 -\frac{108}{a}  = 0\\
3 a^2  +2  b^2 + 3   - \frac{76}{b}   = 0 $$
allows to isolate the terms with $a^2$ and $b^2$ by weighted adding, which gives
$$4 a^2 + 54/a + 3 = 114/b \\
8 b^2 + 38/b + 3 = 162/a$$
Now let $A = 1/a$ and $B = 1/b$ to end with
$$B = \frac{9 A}{19} + \frac{1}{38} + \frac{2}{57 A^2} \\
 A = \frac{19 B}{81} + \frac{1}{54} + \frac{4}{81 B^2}$$
One can make this an iterative pair of equations; start with some $A$, obtain directly $B$ from the first equation, plug into the second and obtain a new $A$, repeat ...
Both equations are of  the same functional form.
Since there must be a solution, the first one should fall faster with $A$ then the second one with $B$, so this iteration should converge.  Here are some values which are obtained when starting with $A=1$:
$$A=1 , B = 0.535, A = 0.3165, B = 0.526, A = 0.320, B = 0.520, A = 0.323, \cdots $$
Convergence is not very fast, but letting  this go for a while leads to values which are approximating $$A=\frac13 , B = \frac12$$ and indeed one can check that this solves the equations.
So the result is $u = 3 + 2 \sqrt{2}$, q.e.d. $\qquad  \Box$
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