If $$x + \frac{1}{x} = 3{\text{,}}$$   where $$x \ne 0{\text{,}}$$  then the value of $$\frac{{{x^4} + 3{x^3} + 5{x^2} + 3x + 1}}{{{x^4} + 1}}$$     = ?

Solution (By Examveda Team)

$$\eqalign{ & x + \frac{1}{x} = 3 \cr & \Rightarrow {x^2} + 1 = 3x\,.....(i) \cr & \Rightarrow {\left( {{x^2} + 1} \right)^2} = {\left( {3x} \right)^2} \cr & \Rightarrow {x^4} + 1 + 2{x^2} = 9{x^2} \cr & \Rightarrow {x^4} + 1 = 7{x^2}\,.....(ii) \cr & \therefore \frac{{{x^4} + 3{x^3} + 5{x^2} + 3x + 1}}{{{x^4} + 1}} \cr & \Rightarrow \frac{{7{x^2} + 3{x^3} + 5{x^2} + 3x}}{{{x^4} + 1}} \cr & \Rightarrow \frac{{12{x^2} + 3{x^3} + 3x}}{{7{x^2}}} \cr & {\text{From equation (i)}} \cr & \Rightarrow \frac{{12x + 3\left( {{x^2} + 1} \right)}}{{7x}} \cr & \Rightarrow \frac{{12x + 3 \times 3x}}{{7x}} \cr & \Rightarrow \frac{{21x}}{{7x}} \cr & \Rightarrow 3 \cr} $$