If $$x+\frac{1}{x}=3$$, then what is the value of $$\frac{x^4+5x^3+3x^2+5x+1}{x^4+1}$$ ?

Given : $$x+\frac{1}{x}=3$$

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

=> $$x^2+1=3x$$ ---------(i)

Squaring both sides, we get :

=> $$(x^2+1)^2=(3x)^2$$

=> $$x^4+1+2x^2=9x^2$$

=> $$x^4+1=9x^2-2x^2=7x^2$$ -----------(ii)

To find : $$\frac{x^4+5x^3+3x^2+5x+1}{x^4+1}$$

= $$\frac{(x^4+1)+5x(x^2+1)+3x^2}{(x^4+1)}$$

Substituting values from equations (i) and (ii),

= $$\frac{7x^2+5x(3x)+3x^2}{7x^2}$$

= $$\frac{25x^2}{7x^2}=\frac{25}{7}$$

=> Ans - (A)