how i can access to red box ${ \frac{x^8-1}{x^4-1} }$ ?? and how complete my solution to end result ?
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2$\begingroup$ What is $\mu(k)$? $\endgroup$– nonuserDec 30, 2019 at 14:48
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2$\begingroup$ I'm not sure what you're asking, but $x^8-1=(x^4)^2-(1)^2=(x^4-1)(x^4+1)$ $\endgroup$– J. W. TannerDec 30, 2019 at 14:48
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3$\begingroup$ @Aqua: cf. Mobius function $\endgroup$– J. W. TannerDec 30, 2019 at 14:51
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2 Answers
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$$(x-1)^{\mu (8)}(x^2-1)^{\mu (4)}(x^4-1)^{\mu (2)}(x^8-1)^{\mu (1)}$$ $$=(x-1)^{0}(x^2-1)^{0}(x^4-1)^{-1}(x^8-1)^{1}$$ $$ {x^8-1\over x^4-1} = {(x^4-1)(x^4+1)\over x^4-1} = x^4+1$$
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$\begingroup$ For the final stage, where you get a fraction with polynomials upstairs and downstairs, one may need to perform polynomial division, as is usually taught in high schools. Here, the final stage is easy, as you demonstrate. $\endgroup$– LubinAug 25, 2020 at 22:58
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Difference of two squares:
$x^8-1=(x^4)^2-(1)^2=(x^4-1)(x^4+1)$,
so $\bbox[white,5px,border:2px solid red] {\dfrac{x^8-1}{x^4-1}}=x^4+1=\Phi_8(x).$