If $ x + \frac{1}{x} = \frac{k}{2}$, then what is the value of $\frac{x^8+1}{x^4}$ ?

$\frac{k^4-16k^2+32}{16}$

If $ x + \frac{1}{x} = \frac{k}{2}$,

then what is the value of $\frac{x^8+1}{x^4}$

We can write $\frac{x^8+1}{x^4}$ as x⁴ + \(\frac{1}{x^4}\)

If $(x + \frac{1}{x}) = x

(x² + x^-2) = (x)² - 2 = b

= (x⁴ + x^-4) = b² - 2

$ x + \frac{1}{x} = \frac{k}{2}$,

(x² + x^-2) = ($ \frac{k}{2}$)² - 2 = \(\frac{k^2 - 8}{4}\)

and, x⁴ + \(\frac{1}{x^4}\)= (\(\frac{k^2 - 8}{4}\))² - 2

x⁴ + \(\frac{1}{x^4}\) = $\frac{k^4-16k^2+32}{16}$