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.

$$x^4 + \frac{1}{x^4} = \frac{257}{16}$$

$$=$$>  $$x^4+\frac{1}{x^4}+2=\frac{257}{16}+2$$

$$=$$>  $$\left(x^2+\frac{1}{x^2}\right)^2=\frac{289}{16}$$

$$=$$>  $$x^2+\frac{1}{x^2}=\frac{17}{4}$$

$$=$$>  $$x^2+\frac{1}{x^2}+2=\frac{17}{4}+2$$

$$=$$>  $$\left(x+\frac{1}{x}\right)^2=\frac{25}{4}$$

$$=$$>  $$x+\frac{1}{x}=\frac{5}{2}$$

$$=$$>  $$\left(x+\frac{1}{x}\right)^3=\left(\frac{5}{2}\right)^3$$

$$=$$>  $$x^3+\frac{1}{x^3}+3.x.\frac{1}{x}\left(x+\frac{1}{x}\right)=\frac{125}{8}$$

$$=$$>  $$x^3+\frac{1}{x^3}+3\left(\frac{5}{2}\right)=\frac{125}{8}$$

$$=$$>  $$x^3+\frac{1}{x^3}=\frac{125}{8}-\frac{15}{2}$$

$$=$$>  $$x^3+\frac{1}{x^3}=\frac{65}{8}$$

$$\therefore\ $$ $$\frac{8}{13}\left(x^3+\frac{1}{x^3}\right)=\frac{8}{13}\times\frac{65}{8}$$

$$=$$>  $$\frac{8}{13}\left(x^3+\frac{1}{x^3}\right)=5$$

Hence, the correct answer is Option D