Finding $x^2 + \frac{1}{x^2}$

We are given the equation $x^4 + \frac{1}{x^4} = 322$. We know the algebraic identity $(a + b)^2 = a^2 + 2ab + b^2$. Let $a = x^2$ and $b = \frac{1}{x^2}$.

Then, $\left(x^2 + \frac{1}{x^2}\right)^2 = (x^2)^2 + 2(x^2)\left(\frac{1}{x^2}\right) + \left(\frac{1}{x^2}\right)^2$

$\left(x^2 + \frac{1}{x^2}\right)^2 = x^4 + 2 + \frac{1}{x^4}$

Substitute the given value: $\left(x^2 + \frac{1}{x^2}\right)^2 = 322 + 2 = 324$.

Taking the square root of both sides: $x^2 + \frac{1}{x^2} = \sqrt{324}$. Since $x > 1$, $x^2 > 0$, thus $x^2 + \frac{1}{x^2}$ must be positive.

$x^2 + \frac{1}{x^2} = 18$.

Finding $x - \frac{1}{x}$

We use the identity $(a - b)^2 = a^2 - 2ab + b^2$. Let $a = x$ and $b = \frac{1}{x}$.

Then, $\left(x - \frac{1}{x}\right)^2 = x^2 - 2(x)\left(\frac{1}{x}\right) + \left(\frac{1}{x}\right)^2$

$\left(x - \frac{1}{x}\right)^2 = x^2 - 2 + \frac{1}{x^2}$

Substitute the value of $x^2 + \frac{1}{x^2}$: $\left(x - \frac{1}{x}\right)^2 = 18 - 2 = 16$.

Taking the square root of both sides: $x - \frac{1}{x} = \sqrt{16}$. Since $x > 1$, we know $x > \frac{1}{x}$, so $x - \frac{1}{x}$ is positive.

$x - \frac{1}{x} = 4$.

Calculating $x^3 - \frac{1}{x^3}$

We use the algebraic identity $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$. Let $a = x$ and $b = \frac{1}{x}$.

$x^3 - \frac{1}{x^3} = \left(x - \frac{1}{x}\right)\left(x^2 + (x)\left(\frac{1}{x}\right) + \left(\frac{1}{x}\right)^2\right)$

$x^3 - \frac{1}{x^3} = \left(x - \frac{1}{x}\right)\left(x^2 + 1 + \frac{1}{x^2}\right)$

Substitute the values we found: $x - \frac{1}{x} = 4$ and $x^2 + \frac{1}{x^2} = 18$.

$x^3 - \frac{1}{x^3} = (4)(18 + 1)$

$x^3 - \frac{1}{x^3} = 4 \times 19 = 76$.