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

5

If x⁴ + \(\frac{1}{x^4}\) = a

then x² + \(\frac{1}{x^2}\) = \(\sqrt {a + 2}\) = b

and x + \(\frac{1}{x}\) = \(\sqrt {b + 2}\)

and we also know that,

If x + \(\frac{1}{x}\)  = n

then, $x^3 +\frac{1}{x^3}$ = n³ - 3 × n

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

then x² + \(\frac{1}{x^2}\) = \(\sqrt {\frac{257}{16} + 2}\) = $\frac{17}{4}$

and x + \(\frac{1}{x}\) = \(\sqrt {\frac{17}{4} + 2}\) = $\frac{5}{2}$

then $(x^3+\frac{1}{x^3})$ = ($\frac{5}{2}$)³ - 3 × $\frac{5}{2}$= $\frac{65}{8}$

So the value of $\frac{8}{13}(x^3+\frac{1}{x^3})$ = $\frac{8}{13}(\frac{65}{8})$ = 5