The problem asks for the value of the expression $\left(x - \frac{1}{x}\right)^3$ given $x = 2 + \sqrt{5}$.

Calculating the Reciprocal of x

First, find the value of $\frac{1}{x}$.

Given $x = 2 + \sqrt{5}$.

Calculate $\frac{1}{x}$: $ \frac{1}{x} = \frac{1}{2 + \sqrt{5}} $ To simplify, multiply the numerator and denominator by the conjugate of the denominator, which is $2 - \sqrt{5}$: $ \frac{1}{x} = \frac{1}{2 + \sqrt{5}} \times \frac{2 - \sqrt{5}}{2 - \sqrt{5}} $ $ \frac{1}{x} = \frac{2 - \sqrt{5}}{(2)^2 - (\sqrt{5})^2} $ $ \frac{1}{x} = \frac{2 - \sqrt{5}}{4 - 5} $ $ \frac{1}{x} = \frac{2 - \sqrt{5}}{-1} $ $ \frac{1}{x} = -(2 - \sqrt{5}) = \sqrt{5} - 2 $

Calculating the Expression $x - \frac{1}{x}$

Now, substitute the values of $x$ and $\frac{1}{x}$ into the expression $x - \frac{1}{x}$.

$ x - \frac{1}{x} = (2 + \sqrt{5}) - (\sqrt{5} - 2) $ $ x - \frac{1}{x} = 2 + \sqrt{5} - \sqrt{5} + 2 $ $ x - \frac{1}{x} = 4 $

Calculating the Final Value

Finally, calculate the cube of the result obtained.

We need to find $\left(x - \frac{1}{x}\right)^3$.

Since $x - \frac{1}{x} = 4$, we have: $ \left(x - \frac{1}{x}\right)^3 = (4)^3 $ $ (4)^3 = 4 \times 4 \times 4 = 64 $

Therefore, the value of $\left(x - \frac{1}{x}\right)^3$ is 64.