If , then the value of is:

Correct option is B

Given:

​We are given $$x + \frac{1}{x} = 7$$​​

Solution:

Square both sides of $$x + \frac{1}{x} = 7 :$$​

​$$\left( x + \frac{1}{x} \right)^2 = 7^2$$​​

​$$x^2 + 2 + \frac{1}{x^2} = 49$$​​

​$$x^2 + \frac{1}{x^2} = 47$$​​

Square again to find $$x^4 + \frac{1}{x^4} :$$​​

​$$\left( x^2 + \frac{1}{x^2} \right)^2 = 47^2$$​​

​$$x^4 + 2 + \frac{1}{x^4} = 2209$$​​

​$$x^4 + \frac{1}{x^4} = 2207$$​​

Use the relation for $$x^6 + \frac{1}{x^6}$$​​

​$$\left( x^2 + \frac{1}{x^2} \right) \left( x^4 + \frac{1}{x^4} \right) = x^6 + \frac{1}{x^6} + x^2 + \frac{1}{x^2}$$​​

​$$47 \times 2207 = x^6 + \frac{1}{x^6} + 47$$​​

​$$103729 = x^6 + \frac{1}{x^6} + 47$$​​

​$$x^6 + \frac{1}{x^6} = 103682$$​​