If $$x^2 + 1 = 3x$$, then the value of $$\frac{(x^4 + x^{-2})}{(x^2 + 5x + 1)}$$ is:

$$x^2 + 1 = 3x$$

$$\frac{1}{x}$$ is multiplied both the sides in the above equation.

$$x+\frac{1}{x}=3$$    Eq.(i)

value of $$\frac{(x^4 + x^{-2})}{(x^2 + 5x + 1)}$$

$$\frac{1}{x}$$ is multiplied in the above equation.

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

$$\frac{(x^3+\frac{1}{x^3})}{(x+5+\frac{1}{x})}$$    Eq.(ii)

take cube of Eq.(i).

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

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

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

Put  Eq.(i) in the above equation.

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

$$x^3+\frac{1}{x^3}+9=27$$

$$x^3+\frac{1}{x^3} = 27-9 = 18$$    Eq.(iii)

Put Eq.(i) and Eq.(iii) in Eq.(ii).

$$\frac{18}{(3+5)}$$

$$\frac{18}{8}$$

$$\frac{9}{4}$$

$$2\frac{1}{4}$$