​If $$x + \frac{1}{x} = 3,$$​  then $$\quad x^6 + \frac{1}{x^6}$$​ is:​

Correct option is D

Given:

​$$x + \frac{1}{x} =3$$​​

to find ; $$x^6 + \frac{1}{x^6}​$$​​

Formula Used: ​

​$$(a+b)^2 = a^2 + b^2 +2ab \\ \ \\ (a+b)^3 = a^3 + b^3 +3ab(a+b)$$ 

$$\left( x^3 + \frac{1}{x^3} \right)^2 = x^6 + 2 + \frac{1}{x^6}$$​​

Solution:

Square both sides of $$x + \frac{1}{x} =3$$​

$$\left( x + \frac{1}{x} \right)^2 = 3^2$$​

$$x^2 + 2 + \frac{1}{x^2} = 9$$​ 

$$x^2 + \frac{1}{x^2} = 9 - 2$$​

$$x^2 + \frac{1}{x^2} = 7$$ 

Cube both sides of $$x + \frac{1}{x} = 3$$  

$$\left( x + \frac{1}{x} \right)^3 = 3^3$$  

$$x^3 + 3x + \frac{3}{x} + \frac{1}{x^3} = 27$$   

$$x^3 + \frac{1}{x^3} + 3 \left( x + \frac{1}{x} \right) = 27$$ 

$$x^3 + \frac{1}{x^3} = 27-9=18$$   

Now 

$$\left( x^3 + \frac{1}{x^3} \right)^2 = x^6 + 2 + \frac{1}{x^6}$$​

Substituting $$x^3 + \frac{1}{x^3} =18:$$ 

$$18^2 = x^6 + 2 + \frac{1}{x^6}$$ 

$$324 - 2 = x^6 + \frac{1}{x^6}$$ 

​$$x^6 + \frac{1}{x^6}= 322$$  

The correct answer is option (d)322.