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.

Note that
$$(x - \frac{1}{x})^3 = x^3 - 3x + \frac{3}{x} - \frac{1}{x^3}$$
$$(x - \frac{1}{x})^3 = x^3 - \frac{1}{x^3} - 3(x - \frac{1}{x})$$
$$x^3 - \frac{1}{x^3} = (x - \frac{1}{x})^3 + 3(x - \frac{1}{x})$$
So if you let $t = x - \frac{1}{x}$ (the value that you're ultimately planning to solve for), then your equation becomes:
$$t^3 + 3t = 108 + 76\sqrt{2}$$
This is a “depressed” cubic equation that you could solve with Cardano's Formula, but that gives you some ugly nested radicals.  So instead, I'm going another route, by finding an equivalent polynomial equation with integer coefficients.
$$(t^3 + 3t - 108)^2 = (76\sqrt{2})^2$$
$$t^6 + 3t^4 - 108t^3 + 3t^4 + 9t^2 - 324t - 108t^3 - 324t + 11664 = 11552$$
$$t^6 + 6t^4 - 216t^3 + 9t^2 - 648t + 112 = 0$$
There's no general formula for a sixth-degree polynomial.  And the Rational Root Theorem fails to provide us with any linear factors.  But it turns out that a quadratic factor exists.
$$(t^2 - 6 t + 1) (t^4 + 6 t^3 + 41 t^2 + 24 t + 112) = 0$$
Solving the quadratic gives $t = 3 \pm 2\sqrt{2}$.  Solving the remaining quartic gives 4 complex solutions:
$$t = -\frac{3}{2} + \sqrt{2} \pm \frac{3}{2} i \sqrt{7 - 4 \sqrt{2}}$$
$$t = -\frac{3}{2} - \sqrt{2} \pm \frac{3}{2} i \sqrt{7 + 4 \sqrt{2}}$$
But of these 6 roots, it turns out that only one of them is a valid solution to the original equation.
$$t = \boxed{3 + 2\sqrt{2}}$$
