If $$x^4+\frac{1}{x^4}=322$$​, then $$x^3-\frac{1}{x^3}$$​ = ?

Correct option is B

Given:

$$x^4 + \frac{1}{x^4} = 322$$​​
We need to find the value of:
​$$x^3 - \frac{1}{x^3}$$​​
Concept Used:
Algebraic Identities:
​$$\left( x+\frac{1}{x} \right) ^2 = x^2 + \frac{1}{x^2} + 2$$​​
​$$\left( x-\frac{1}{x} \right) ^2 = x^2 + \frac{1}{x^2} - 2$$​​
​$$x^3 - \frac{1}{x^3} = \left(x - \frac{1}{x}\right)\left(x^2 + 1 + \frac{1}{x^2}\right)$$​​
Solution:
​$$\left(x^2 + \frac{1}{x^2}\right)^2 = x^4 + \frac{1}{x^4} + 2 = 322 + 2 = 324$$​
​$$x^2 + \frac{1}{x^2} = \sqrt{324} = 18$$​
​$$\left(x - \frac{1}{x}\right)^2 = x^2 + \frac{1}{x^2} - 2 = 18 - 2 = 16$$​
​$$x - \frac{1}{x} = \sqrt{16} = 4$$​
​$$x^3 - \frac{1}{x^3} = \left(x - \frac{1}{x}\right)\left(x^2 + 1 + \frac{1}{x^2}\right)$$​
​$$x^3 - \frac{1}{x^3} = 4 \times (18 + 1) = 4 \times 19 = 76$$​