If $${x^2} + \frac{1}{{25{x^2}}} = \frac{8}{5}$$   and x > 0, then what is the value of $${x^3} + \frac{1}{{125{x^3}}} = ?$$

Solution(By Examveda Team)

$$\eqalign{ & \because \,{x^2} + \frac{1}{{25{x^2}}} = \frac{8}{5} \cr & \Rightarrow {x^2} + \frac{1}{{25{x^2}}} + 2x \times \frac{1}{{5x}} = \frac{8}{5} + \frac{2}{5} \cr & \Rightarrow {\left( {x + \frac{1}{{5x}}} \right)^2} = 2 \cr & \Rightarrow x + \frac{1}{{5x}} = \sqrt 2 \cr & {\text{On cubing both sides}} \cr & \Rightarrow {x^3} + \frac{1}{{125{x^3}}} + 3 \times \frac{1}{5}\left( {\sqrt 2 } \right) = {\left( {\sqrt 2 } \right)^3} \cr & \Rightarrow {x^3} + \frac{1}{{125{x^3}}} = 2\sqrt 2 - \frac{{3\sqrt 2 }}{5} \cr & \Rightarrow {x^3} + \frac{1}{{125{x^3}}} = \frac{{7\sqrt 2 }}{5} \cr} $$