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

Given,  $$x + \frac{1}{x} = 4$$

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

$$=$$>  $$x^2+\frac{1}{x^2}+2.x.\frac{\ 1}{x}=16$$

$$=$$>  $$x^2+\frac{1}{x^2}+2=16$$

$$=$$>  $$x^2+\frac{1}{x^2}=14$$

$$=$$>  $$\left(x^2+\frac{1}{x^2}\right)^2=14^2$$

$$=$$>  $$x^4+\frac{1}{x^4}+2.x^2.\frac{\ 1}{x^2}=196$$

$$=$$>  $$x^4+\frac{1}{x^4}+2=196$$

$$=$$>  $$x^4+\frac{1}{x^4}=196-2$$

$$=$$>  $$x^4+\left(\frac{1}{x}\right)^4=194$$

Hence, the correct answer is Option C