<s>[INST] 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?
 [/INST] You can still factor this. Notice that 
\begin{align*}
x^5+x^4+x^3+x^2+x+1 &= x^3(x^2+x+1)+1(x^2+x+1) \\
&=(x^3+1)(x^2+x+1) \\
&=(x+1)(x^2-x+1)(x^2+x+1).
\end{align*}
Now you can go one step further and show that both $(x^2-x+1)$ and $(x^2+x+1)$ are irreducible in the real numbers just by using the discriminant. 
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