How to factorize :$f(x)=x^n+x+1 \ \ \ \ \ \ : n=3k+2 ,k\in \mathbb{N}$

Question

How to factorize :

$$f(x)=x^n+x+1 \ \ \ \ \ \ : n=3k+2 ,k\in \mathbb{N}$$

And :

$$g(x)=x^n+x-1 \ \ \ \ \ \ : n=3k+2 ,k\in2m-1 \ \ \ , \ \ m\in\mathbb{N}$$

My try :

$$f(x)=x^n+x+1=x^{3k+2}+x+1$$

$$=(x^{3k+2}+x^{3k+1}+x^{3k})-(x^{3k+2}+x^{3k+1}+x^{3k})+(x+1)$$

Now what ?

Answer 1

$$x^{3k+2}+x+1=x^{3k+2}-x^2+x^2+x+1=x^2((x^3)^k-1)+x^2+x+1$$ and use $x^3-1=(x-1)(x^2+x+1)$.

For $n=6m-1$ we have $$x^n+x-1=x^n+x^2-x^2+x-1=x^2(x^{n-2}+1)-(x^2-x+1)$$ and use $x^3+1=(x+1)(x^2-x+1).$

Answer 2

Hint: evaluate $f(x)$ at $\omega$, where $\omega=e^{2\pi i/3}$, so $\omega^3=1$, and $\omega^2 + \omega + 1 = 0$.