<s>[INST] Is $x^6 + 1$ divisible by $x^2 + 1$? I used polynomial long division and found that the remainder was $x + 1$.  So I concluded that $x^6 + 1$ is not divisble by $x^2 + 1$.  But this is not true since $x^6 + 1$ can be factored as a sum of cubes of $x^2$ and $1$.
What can we conclude if we use long division on two polynomials and get a non-zero remainder? 
 [/INST] $$x^6+1=(x^2+1)(x^4-x^2+1)$$
as à consequence of $a^3+b^3=(a+b)(a^2-ab+b^2)$
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