Factoring $x^5 + x^4 + x^3 + x^2 + x + 1$ without using $\frac{x^n - 1}{x-1}$? I was at a math team meet today and one of the problems was to factor $x^5 + x^4 + x^3 + x^2 + x + 1$. It also gave the hint that it decomposes into two trinomials and a binomial.
The solution they gave was based on the fact that $\frac{x^6 - 1}{x-1} = x^5 + x^4 + x^3 + x^2 + x + 1$ and from there the solution is pretty straightforward. However, I was not aware of that factorization. The only ones I have really learned are $x^2 - y^2 = (x-y)(x+y)$ and $x^3 \pm y^3 = (x \pm y)(x^2 \mp xy + y^2)$. Is there any other way I could have solved this factorization without using the ones they used?

Here's how you could have found that identity without having known it before hand.
First, notice that setting $x = -1$ gives $0$. Thus, $(x - (-1)) = x+1$ is a root.
Using polynomial long division, we find that we can reduce it to $(x^4 + x^2 + 1)(x+1)$
Now, what to do with the $x^4 + x^2 + 1$ term? Recall that $x^3 - 1 = (x-1)(x^2 + x + 1)$. If we plug $x = x^2$ into this identity, we find $x^6 - 1 = (x^2 - 1)(x^4 + x^2 + 1)$, so $x^4 + x^2 + 1 = \frac{x^6 - 1}{x^2 - 1}$.
We have now reduced it to $\frac{(x^6 - 1)(x+1)}{(x-1)(x+1)}$ = $\frac{x^6 - 1}{x-1}$ and, as you said, the rest is easy from here using the identities you listed in the OP.
