If $x^2+\frac{1}{x^2}= 18,x > 0$, then find the value of $x^3+\frac{1}{x^3}$.

$34\sqrt{5}$

If $(x^2+\frac{1}{x^2}) = 18, x > 0 $

$(x^3+\frac{1}{x^3})$= ?

If $K^2+\frac{1}{K^2}$ = n 

then,  K+ \(\frac{1}{k}\) = \(\sqrt {n + 2}\)

x+ \(\frac{1}{x}\) = \(\sqrt {18 + 2}\) = 2\(\sqrt {5}\) 

$(x^3+\frac{1}{x^3})$= (2\(\sqrt {5}\)) ³ - 3 × 2\(\sqrt {5}\)  = $40\sqrt{5}$ - $6\sqrt{5}$

= $(x^3+\frac{1}{x^3})$ = $34\sqrt{5}$