2016/063) Prove for all integers $N>1$ $(N^2)^{2014}- (N^{11})^{106}$ is divisible by $N^6+ N^3+1$

$(N^2)^{2014}- (N^{11})^{106}$
$=N^{4028} - N^{1166}$
$=N^{1166}(N^{2862}-1) = N^{1166}((N^9)^{318}-1)$ is divisible by $N^9-1$
as $N^9-1 = (N^3)^3 -1 = (N^3-1)(N^6+N^3+1)$
as $(N^2)^{2014}- (N^{11})^{106}$ is divsible by $N^9-1$ which is divisible by $N^6+N^3+1$ hence  $(N^2)^{2014}- (N^{11})^{106}$
is divisible by $N^6+N^3+1$