If $$x-\frac{1}{x}=5$$​, find the value of $$x^4+\frac{1}{x^4}$$​.

Correct option is B

Given:
​$$x-\frac{1}{x}=5$$ 
Formula Used:
​$$(x - \frac{1}{x})^2 = x^2 + \frac{1}{x^2}- 2\\\ \\(x + \frac{1}{x})^2 = x^2 + \frac{1}{x^2} + 2$$​​
Solution:
​$$x-\frac{1}{x}=5$$ 
Squaring both sides
​$$(x-\frac{1}{x})^2=25\\x^2 + \frac{1}{x^2} - 2 = 25 \\x^2 + \frac{1}{x^2}= 27$$ 
Again Squaring both sides 
 $$(x^2 + \frac{1}{x^2})^2= 729\\x^4 + \frac{1}{x^4} = 729-2 = 727\\$$​​