$\omega^2+\omega+1$divides a polynomial The question is 
Show that $f(n)=n^5+n^4+1$ is not prime for $n>4$.
The solution is given as
Let $\omega$ be the third root of unity. Then $\omega^2+\omega+1=0$. Since $\omega^5+\omega^4+1=\omega^2+\omega+1$, we see that $\omega^2+\omega+1$ is a *factor of the polynomial. So *$n^2+n+1|n^5+n^4+1$.
Which polynomial are we referring to in the bold typeface above? And how is $n^2+n+1|n^5+n^4+1$ true due to $\omega^5+\omega^4+1=\omega^2+\omega+1$?

Let $$g(n)=n^2+n+1$$ and $$h(n)=n^3-n+1$$.
Clearly $$f(n)=g(n) \cdot h(n)$$
Now $h(1)=1$ and $h'(n) = 3 n^2-1 > 0$ for $n\ge 1$. So, for $n>1$ both $g(n)$ and $h(n)$ are greater than $1$. So
$f(n)$ **is composite for all $n>1$
