If $${x^4} + \frac{1}{{{x^4}}} = \frac{{257}}{{16}},$$   then find $$\frac{8}{{13}}\left( {{x^3} + \frac{1}{{{x^3}}}} \right),$$   where x > 0.

Solution(By Examveda Team)

$$\eqalign{ & {x^4} + \frac{1}{{{x^4}}} = \frac{{257}}{{16}} \cr & {\left( {{x^2} + \frac{1}{{{x^2}}}} \right)^2} = \frac{{257}}{{16}} + 2 \cr & {\left( {{x^2} + \frac{1}{{{x^2}}}} \right)^2} = {\left( {\frac{{17}}{4}} \right)^2} \cr & {x^2} + \frac{1}{{{x^2}}} = \frac{{17}}{4} \cr & x + \frac{1}{x} = {\left( {\frac{{17}}{4} + 2} \right)^{\frac{1}{2}}} = \frac{5}{2} \cr & {\text{Here }}x = 2 \cr & {\text{Hence }}\frac{8}{{13}}\left( {{x^3} + \frac{1}{{{x^3}}}} \right) \cr & = \frac{8}{{13}}\left( {8 + \frac{1}{8}} \right) \cr & = \frac{8}{{13}} \times \frac{{65}}{8} \cr & = 5 \cr} $$