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.
$$x^4 + \frac{1}{x^4} = \frac{257}{16}$$
$$=$$> Â $$x^4+\frac{1}{x^4}+2=\frac{257}{16}+2$$
$$=$$> Â $$\left(x^2+\frac{1}{x^2}\right)^2=\frac{289}{16}$$
$$=$$> Â $$x^2+\frac{1}{x^2}=\frac{17}{4}$$
$$=$$> Â $$x^2+\frac{1}{x^2}+2=\frac{17}{4}+2$$
$$=$$> Â $$\left(x+\frac{1}{x}\right)^2=\frac{25}{4}$$
$$=$$> Â $$x+\frac{1}{x}=\frac{5}{2}$$
$$=$$> Â $$\left(x+\frac{1}{x}\right)^3=\left(\frac{5}{2}\right)^3$$
$$=$$> Â $$x^3+\frac{1}{x^3}+3.x.\frac{1}{x}\left(x+\frac{1}{x}\right)=\frac{125}{8}$$
$$=$$> Â $$x^3+\frac{1}{x^3}+3\left(\frac{5}{2}\right)=\frac{125}{8}$$
$$=$$> Â $$x^3+\frac{1}{x^3}=\frac{125}{8}-\frac{15}{2}$$
$$=$$> Â $$x^3+\frac{1}{x^3}=\frac{65}{8}$$
$$\therefore\ $$ $$\frac{8}{13}\left(x^3+\frac{1}{x^3}\right)=\frac{8}{13}\times\frac{65}{8}$$
$$=$$> Â $$\frac{8}{13}\left(x^3+\frac{1}{x^3}\right)=5$$
Hence, the correct answer is Option D
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