Maybe by checking the complexe roots of $N^6+N^3+1$ and see that they are also roots of the other expression

It is even divisible by $N^9-1=(N^3-1)(N^6+N^3+1)$. Just note that the original term is $N^{1166} \cdot (N^{2862}-1)$ and $9 \mid 2862$.

We know that the first expression can be written as $N^{1166}(N^{2862}-1)$. also note that $N^6 + N^3 +1$ is the 9th cyclotomic polynomial. Since $9 \mid 2862$ and by properties of cyclotomics, we conclude the result. EDIT: ive been sniped

Thanks guys!

Prove that for $N>1$ that $(N^{2})^{2014} - (N^{11})^{106}$ is divisible by $N^6 + N^3 +1$