If $$x^2+\frac{1}{x^2}=14$$​ then what is the value of $$x^3+\frac{1}{x^3} ?(\mathrm{x}>0)$$

Correct option is B

Given:

$$x^2 + \frac{1}{x^2} = 14,\quad x > 0 \\\text{Find: } x^3 + \frac{1}{x^3} = ?$$

Formula:

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

Solution:

$$x^2 + \frac{1}{x^2} = 14 \Rightarrow \left(x + \frac{1}{x}\right)^2 = 14 + 2 = 16 \\\Rightarrow x + \frac{1}{x} = \sqrt{16} = 4 \\\ \\x^3 + \frac{1}{x^3} = \left(x + \frac{1}{x}\right)^3 - 3\left(x + \frac{1}{x}\right) \\\ \\= 4^3 - 3 \times 4 = 64 - 12 = \boxed{52}$$

​Final Answer: (B) 52