If $$x + \frac{1}{x} = \sqrt5$$. then $$x^3 + \frac{1}{x^3}$$ is equal to:

$$x + \frac{1}{x} = \sqrt5$$

Apply this formula $$(a+b)^3$$ = $$a^3 + b^3+ 3 × a × b (a+b)$$

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

$$\sqrt5^3$$ = $$x^3 + \frac{1}{x^3}$$ + 3 × ($$x + \frac{1}{x}$$)

$$\sqrt5^3$$ - $$3 \sqrt5$$ =  $$x^3 + \frac{1}{x^3}$$

Hence $$x^3 + \frac{1}{x^3}$$ = $$5\sqrt25$$ - $$3 \sqrt5$$

 $$x^3 + \frac{1}{x^3}$$ = $$5\sqrt5$$ -  $$3 \sqrt5$$

Therefore $$x^3 + \frac{1}{x^3}$$ = $$2 \sqrt5$$