If $x^{2}-3x+1=0 $, then the value of $ (x^{4}+\frac{1}{x^{2}}) \div(x^{2}+1)$ is :

6

If x + \(\frac{1}{x}\)  = n

then, $x^3 +\frac{1}{x^3}$ = n³ - 3 × n

If $x^{2}-3x+1=0 $

then the value of $\left(x^{4} + \frac{1}{x^{2}}\right) \div (x^{2} + 1)$ = ?

we can write $ (x^{4}+\frac{1}{x^{2}}) \div(x^{2}+1)$ by taking x as common form numerator and denominator as = 

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

$x^{2}- 3x + 1 = 0$

Divide by x on both sides,

x + \(\frac{1}{x}\) = 3

then, $x^3 +\frac{1}{x^3}$ = 3³ - 3 × 3 = 18

So the value of $\left(x^{3} + \frac{1}{x^{3}}\right) \div (x + \frac{1}{x})$ = \(\frac{18}{3}\) = 6