\[ \text { If } x^4+\frac{1}{x^4}=322 \text {, and } x>1 \text { then the value of } x^3-\frac{1}{x^3} \text { is } \]

Given:
– \(x^4 + \frac{1}{x^4} = 322\) and \(x > 1\).

1. We know that \((a + b)^2 = a^2 + b^2 + 2ab\). Therefore, \((a + b)^2 – 2ab = a^2 + b^2\).

2. Applying this to \(x^2 + \frac{1}{x^2}\):
\[ \left(x^2 + \frac{1}{x^2}\right)^2 – 2 \times x^2 \times \frac{1}{x^2} = x^4 + \frac{1}{x^4} \]
\[ \left(x^2 + \frac{1}{x^2}\right)^2 = 322 + 2 \]
\[ \left(x^2 + \frac{1}{x^2}\right)^2 = 324 \]
\[ x^2 + \frac{1}{x^2} = \pm 18 \]
Since \(x > 1\), we take the positive root:
\[ x^2 + \frac{1}{x^2} = 18 \]

3. Similarly, for \(x^3 – \frac{1}{x^3}\), we use the identity \((a – b)^2 + 2ab = a^2 + b^2\):
\[ \left(x^2 – \frac{1}{x^2}\right)^2 + 2 = 18 \]
\[ \left(x^2 – \frac{1}{x^2}\right)^2 = 16 \]
\[ x^2 – \frac{1}{x^2} = \pm 4 \]
Since \(x > 1\), we take the positive root:
\[ x^2 – \frac{1}{x^2} = 4 \]

4. Now, cubing both sides:
\[ \left(x^3 – \frac{1}{x^3}\right) – 3 \times x \times \frac{1}{x} \left(x^2 – \frac{1}{x^2}\right) = 64 \]
\[ x^3 – \frac{1}{x^3} – 3 \times 4 = 64 \]
\[ x^3 – \frac{1}{x^3} = 64 + 12 \]
\[ x^3 – \frac{1}{x^3} = 76 \]

Conclusion:
The value of \(x^3 – \frac{1}{x^3}\) is 76.