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

Given $$x+\frac{1}{x} = 5$$ or $$x^{2} - 5x +1 = 0$$
And we have to find value of $$\frac{x^4 + \frac{1}{x^2}}{x^2 - 3x +1}$$

or $$\frac{x^4 + \frac{1}{x^2}}{x^2 - 5x +1 + 2x}$$

or $$\frac{x^4 + \frac{1}{x^2}}{2x}$$

or $$\frac{x^3 + \frac{1}{x^3}}{2}$$

( As $$(x+\frac{1}{x}) = 5$$ hence $$(x+\frac{1}{x})^3 = 125$$ or $$x^3 + \frac{1}{x^3} = 110$$ )

So now $$\frac{x^3 + \frac{1}{x^3}}{2}$$ = 110/2 = 55