<s>[INST] Polynomial Divisibilty Test I recently came across a question in a book and I was wondering how to go about solving this. I just need a hint about how I could approach it.  
I have to show that $x^{6n+2} - x^{6n+1} + 1$  is always divisible by $x^2 - x + 1$ where $n = \{1,2,3,4,\cdots\}$  
 [/INST] Let $\alpha,\beta$ be two (complex) roots of $x^2-x+1$. Then $x^2-x+1=(x-\alpha)(x-\beta)$. Note that $x^2-x+1$ divides $x^3+1=(x+1)(x^2-x+1)$, and hence also $x^6-1=(x^3-1)(x^3+1)$. So both $\alpha,\beta$ satisfy $x^6=1$. Thus we have $\alpha^{6n+2}-\alpha^{6n+1}+1=(\alpha^6)^n\cdot\alpha^2-(\alpha^6)^n\cdot\alpha+1=\alpha^2-\alpha+1=0$, and same for $\beta$. So both $\alpha,\beta$ are roots of $x^{6n+2}-x^{6n+1}+1$. So $x-\alpha,x-\beta$ divide $x^{6n+2}-x^{6n+1}+1$, so also $(x-\alpha)(x-\beta)=x^2-x+1$ divides $x^{6n+2}-x^{6n+1}+1$.
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