If $x+\frac{1}{x}=2 \sqrt{5}$, then what is the value of $\frac{\left(x^4+\frac{1}{x^2}\right)}{x^2+1} ?$

17

If x + \(\frac{1}{x}\)  = n

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

If $x+\frac{1}{x}=2 \sqrt{5}$

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

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

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

then, $x^3 +\frac{1}{x^3}$ = $(2 \sqrt{5}$)³ - 3 × $2 \sqrt{5}$ = $34 \sqrt{5}$

So the value of $\left(x^{3} + \frac{1}{x^{3}}\right) \div (x + \frac{1}{x})$ = \(\frac{34 \sqrt{5}}{2 \sqrt{5}}\) = 17