If x² + $$\frac 1{x^2}$$ =18 , and x>0, what is the value of x³ + $$\frac1{x^3}$$​​

Correct option is D

​$$x^2 + \frac{1}{x^2} = 18,\quad x>0$$​​
Find
​$$x^3 + \frac{1}{x^3}$$​​

Formula Used :
​$$x^2 + \frac{1}{x^2} = \left(x + \frac{1}{x}\right)^2 - 2$$​​
​$$x^3 + \frac{1}{x^3} = \left(x + \frac{1}{x}\right)^3 - 3\left(x + \frac{1}{x}\right)$$​​

Solution :

​$$x + \frac{1}{x} = \sqrt{18 + 2} = \sqrt{20} = 2\sqrt{5}$$​​
(since x > 0)

$$x^3 + \frac{1}{x^3} \\ \ \\= (2\sqrt{5})^3 - 3(2\sqrt{5})\\ \ \\= 8(5\sqrt{5}) - 6\sqrt{5}\\ \ \\= 40\sqrt{5} - 6\sqrt{5}\\ \ \\= 34\sqrt{5}$$​