<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] I thought it might be helpful with answering now.
\begin{align}
\newcommand{w}{\omega}
& x^3-\frac{1}{x^3}=108+76\sqrt{2}. \\
& 108+76\sqrt{2} = \left(x-\frac 1 x\right)\left(x^2+1+\frac 1 {x^2}\right) \\
& = \left(x-\frac 1 x \right)^3+3x\cdot \frac 1 x \left(x-\frac 1 x\right). \\
\ \\
& u^3+3u=108+76\sqrt{2}. \\
& 108+76\sqrt{2}= 4(27+19\sqrt{2})=4(5+3\sqrt{2})(1+\sqrt{2})^2. \\
& (5+3\sqrt{2})^2+3=46+30\sqrt{2} \neq 12+8\sqrt{2}. \\
& \vdots \\
& (3+2\sqrt{2})^2+3=20+12\sqrt{2}=4(5+3\sqrt{2}). \\
\therefore \; & u=3+2\sqrt{2}.
\end{align}
Substituting $x-\dfrac 1 x=3+2\sqrt{2}$, it works.
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