If $$x+\frac{1}{x}=3$$ then $$x^{8}+\frac{1}{x^8}$$ is equal to 

Expression : $$x+\frac{1}{x}=3$$

Squaring both sides,

=> $$(x+\frac{1}{x})^2=(3)^2$$

=> $$x^2+\frac{1}{x^2}+2.x.\frac{1}{x}=9$$

=> $$x^2+\frac{1}{x}^2=9-2=7$$

Squaring both sides, we get : 

=> $$(x^2+\frac{1}{x^2})^2=(7)^2$$

=> $$x^4+\frac{1}{x^4}+2.x^2.\frac{1}{x^2}=49$$

=> $$x^4+\frac{1}{x}^4=49-2=47$$

Again squaring both sides, 

=> $$(x^4+\frac{1}{x^4})^2=(47)^2$$

=> $$x^8+\frac{1}{x^8}+2.x^4.\frac{1}{x^4}=2209$$

=> $$x^8+\frac{1}{x}^8=2209-2=2207$$

=> Ans - (C)