​$$\text{If } x - \frac{1}{x} = 3, \text{ then the value of } x^4 + \frac{1}{x^4} \text{ is:}$$​​

Correct option is D

Given:

​$$x-\dfrac{1}{x}=3$$​​

Formula Used:

​$$(x-\dfrac{1}{x})^2=x^2+\dfrac{1}{x^2}-2$$​​

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

Solution:
​$$(x-\dfrac{1}{x})^2=3^2=9 \implies x^2+\dfrac{1}{x^2}=9+2=11$$​​
​$$\left(x^2+\dfrac{1}{x^2}\right)^2=11^2=121= x^4+\dfrac{1}{x^4}+2$$​​
​$$x^4+\dfrac{1}{x^4}=121-2=119$$​​