If $x^2-5 \sqrt{2} x+1=0$, then what is the value of $\frac{\left(x^3+\frac{1}{x}\right)}{x^2+1} ?$

$\frac{24 \sqrt{2}}{5}$

If $K+\frac{1}{K}=n$

then, $K^2+\frac{1}{K^2}$ = n² – 2

If $x^2-5 \sqrt{2} x+1=0$

Divide by x we get,

x + \(\frac{1}{x}\) = $5 \sqrt{2}$

$x^2 +\frac{1}{x^2}$ = ($5 \sqrt{2}$)² - 2 = 50 - 2 = 48

$\frac{\left(x^3+\frac{1}{x}\right)}{x^2+1}$ = Take x as common 

= $\frac{\left(x^2+\frac{1}{x^2}\right)}{x+\frac{1}{x}\right}$ = ${48}{5 \sqrt{2}}$ 

= $\frac{24 \sqrt{2}}{5}$