If $${x^2} + \frac{1}{{{x^2}}} = \frac{{31}}{9}$$   and x > 0, then what is the value of $${x^3} + \frac{1}{{{x^3}}} = ?$$

Solution(By Examveda Team)

$$\eqalign{ & {x^2} + \frac{1}{{{x^2}}} = \frac{{31}}{9} \cr & x + \frac{1}{x} = \sqrt {\frac{{31}}{9} + 2} \cr & x + \frac{1}{x} = \frac{7}{3} \cr & {x^3} + \frac{1}{{{x^3}}} = {n^3} - 3n \cr & {x^3} + \frac{1}{{{x^3}}} = {\left( {\frac{7}{3}} \right)^3} - 3 \times \frac{7}{3} \cr & {x^3} + \frac{1}{{{x^3}}} = \frac{{343}}{{27}} - 7 \cr & {x^3} + \frac{1}{{{x^3}}} = \frac{{343 - 7 \times 27}}{{27}} \cr & {x^3} + \frac{1}{{{x^3}}} = \frac{{154}}{{27}} \cr} $$