If $${x^2} + \frac{1}{{{x^2}}} = \frac{7}{4}$$   for x > 0 then what is the value of $${x^3} + \frac{1}{{{x^3}}} = ?$$

Solution(By Examveda Team)

$$\eqalign{ & {\text{Given, }}{x^2} + \frac{1}{{{x^2}}} = \frac{7}{4} \cr & {\text{Adding 2 to both sides}} \cr & {x^2} + \frac{1}{{{x^2}}} + 2 = \frac{7}{4} + 2 \cr & {\left( {x + \frac{1}{x}} \right)^2} = \frac{{15}}{4} \cr & x + \frac{1}{x} = \frac{{\sqrt {15} }}{2} \cr & \because \,{x^3} + \frac{1}{{{x^3}}} = {\left( {x + \frac{1}{x}} \right)^3} - 3\left( {x + \frac{1}{x}} \right) \cr & \therefore \,{x^3} + \frac{1}{{{x^3}}} = \frac{{15 \times \sqrt {15} }}{8} - 3 \times \frac{{\sqrt {15} }}{2} \cr & = \frac{{15\sqrt {15} - 12\sqrt {15} }}{8} \cr & = \frac{{3\sqrt {15} }}{8} \cr} $$