If $$x+\frac{1}{x}=17$$, then what is the value of $$\frac{x^4+\frac{1}{x^2}}{x^2-3x+1}$$ ?

Given : $$x+\frac{1}{x}=17$$ ------------(i)

=> $$\frac{x^2+1}{x}=17$$

=> $$x^2+1=17x$$ --------(ii)

Cubing equation (i), we get :

=> $$(x+\frac{1}{x})^3=(17)^3$$

=> $$x^3+\frac{1}{x^3}+3(x)(\frac{1}{x})(x+\frac{1}{x})=4913$$

=> $$x^3+\frac{1}{x^3}+3(17)=4913$$

=> $$x^3+\frac{1}{x^3}=4913-51=4862$$ -------------(iii)

To find : $$\frac{x^4+\frac{1}{x^2}}{x^2-3x+1}$$

Substituting value from equation (ii),

= $$\frac{x^4+\frac{1}{x^2}}{17x-3x}=\frac{x^4+\frac{1}{x^2}}{14x}$$

= $$\frac{1}{14}(x^3+\frac{1}{x^3})$$

Substituting value from equation (iii),

= $$\frac{4862}{14}=\frac{2431}{7}$$

=> Ans - (A)