how solve Cyclotomic Polynomial , n=8

Question

how i can access to red box ${ \frac{x^8-1}{x^4-1} }$ ?? and how complete my solution to end result ?

Answer 1

$$(x-1)^{\mu (8)}(x^2-1)^{\mu (4)}(x^4-1)^{\mu (2)}(x^8-1)^{\mu (1)}$$ $$=(x-1)^{0}(x^2-1)^{0}(x^4-1)^{-1}(x^8-1)^{1}$$ $$ {x^8-1\over x^4-1} = {(x^4-1)(x^4+1)\over x^4-1} = x^4+1$$

Answer 2

Difference of two squares:

$x^8-1=(x^4)^2-(1)^2=(x^4-1)(x^4+1)$,

so $\bbox[white,5px,border:2px solid red] {\dfrac{x^8-1}{x^4-1}}=x^4+1=\Phi_8(x).$