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.


$u^3+3u=108+76\sqrt{2}$

A few posts here assumed $u = a + b\sqrt{2}$, then search for rational $(a, b)$
Let "slope", $\displaystyle \;m_1 = b/a$
$\displaystyle u^2 = (a^2+2b^2) \;+\; (2ab)\,\sqrt{2}\qquad\; →m_2 = \frac{2\,m_1}{1+2\,m_1^2}$
Note that $m_2$ has same sign as $\displaystyle m_1 \quad →m_∞ = \frac{sgn(m1)}{\sqrt{2}}$
Slope of RHS = $\displaystyle \frac{76}{108} ≈ 0.7037 
\;<\; \frac{1}{\sqrt{2}} ≈ 0.7071$
$u^3 ≈ RHS,\;m_1$ likely convergent of RHS slope.
C:\>spigot -C 76/108
0/1
1/1
2/3
5/7
7/10
19/27

Convergent 2/3 work! Upon checking, $(b,a) = (2,3)$, not just the ratio.
$u^3 + 3u = (99+70\sqrt{2}) + 3×(3+2\sqrt{2}) = 108+76\sqrt{2}$
