show that $11\nmid n^6-2n^5+4n^4-8n^3+16n^2-32n+64 $ Let $n$ be postive integer,show that
  $$11\nmid n^6-2n^5+4n^4-8n^3+16n^2-32n+64 $$


I have use ugly metods to solve consider this case,$n\equiv 0,\pm 1,\pm 2,\pm 3,\pm 4,\pm 5\pmod {11}$,have other more simple methods it?
Starting from @Crostul's comment, we have
$$
f(n)=n^6-2n^5+4n^4-8n^3+16n^2-32n+64 = \frac{n^7+2^7}{n+2}
$$
and so $f(n)(n+2)=n^7+2^7$.
Therefore, if $11$ divides $f(n)$, then $11$ divides $n^7+2^7$.

Now $n \mapsto n^7$ is an injective map mod $11$ because $\gcd(7,\phi(11))=1$.

Therefore, if $11$ divides $f(n)$, then $n^7 \equiv -2^7 = (-2)^7 \bmod 11$ and so $n \equiv -2 \bmod 11$.

So you only have to check that $11$ does not divide $f(n)$ for $n=-2$.
Guess $11|n^6−2n^5+4n^4−8n^3+16n^2−32n+64$, then $11|n^7+2^7$ so $n^7 \equiv-2^7 \pmod {11}.$ By Euler's theorem, $n^{10} \equiv 1 \pmod {11}$, so $n \Rightarrow n^{21} \Rightarrow (n^7) ^3 \Rightarrow -2^{21} \Rightarrow \equiv -2 \pmod {11} .$ By distributing the exponent, we have $11v(n^7+2^7) \Rightarrow 11v(n+2)+11v(7) \Rightarrow 11(n+2)$, so $(n^6−2n^5+4n^4−8n^3+16n^2−32n+64) = 11v(n^7+2^7)-11v(n+2) = 0$.