Factoristaion of $x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1$ My question revolves around the polynomial
$$x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1$$
I know that it can be factorised into
$$(x^2+x+1)(x^6+x^3+1)$$
but what method can we employ to obtain this result in the first place? Any method that works will be of great help to me.
Thank you for your help.
 A: Method 1: Use what you know about polynomials of the form $x^n-1$. For instance, your polynomial is $\frac{x^9-1}{x-1}$. So if you can factor the numerator, many of those factors would be inherited to your polynomial. And indeed, we have
$$
x^9-1=(x^3)^3-1=((x^3)^2+x^3+1)(x^3-1)\\
=(x^6+x^3+1)(x^2+x+1)(x-1)
$$
Put this into the fraction above, and the factorisation follows.
Method 2: Inspection. There are nine terms. They can be grouped three-by-three:
$$
(x^8+x^7+x^6)+(x^5+x^4+x^3)+(x^2+x+1)\\
=x^6(x^2+x+1)+x^3(x^2+x+1)+(x^2+x+1)
$$
and the factorisation follows.
A: The essential key to answering your question is to recognize that your polynomial is $(x^9-1)\big/(x-1)$.
Thus, in effect, you’re asking for the complete factorization into $\Bbb Z$-irreducible factors, of $x^9-1$. @Arthur, who gave the excellent answer that you accepted, undoubtedly knows all I’m about to say, but he kindly and reticently decided not to overwhelm you. I, being neither kind nor reticent, am about to charge ahead and attempt to give you a comprehensible explanation of (at least some of) the big picture.
The complete factorization of $x^9-1$ is $\Phi_1(x)\Phi_3(x)\Phi_9(x)$, where the Phi’s are cyclotomic polynomials: $\Phi_d(x)$ is the (necessarily integral) polynomial whose roots are all the primitive $d$-th roots of unity in $\Bbb C$. These all are $\Bbb Z$-irreducible: easy to show irreducibility when $d$ is a prime power, but requiring a rather more advanced proof otherwise.
The complete factorization of $x^n-1$ is:
$$
x^n-1=\prod_{d|n}\Phi_d(x)\,,
$$
and I’ll justify this way further on, perhaps long after you’ve quit reading. (“tl;dnr”)
But if you’ll accept this for now, I’ll turn the displayed formula into a method of writing down these mysterious $\Phi_d$’s.
It involves a multiplicative application of the famous Möbius Inversion Formula. The standard additive version says that if $F$ and $G$ are functions defined on the positive integers (with value in any given abelian group) such that
$$
\text{If}\quad(\forall n\ge1)\bigl(G(n)=\sum_{d|n}F(d)\bigr)\quad\text{then}\quad
(\forall n\ge1)\bigl(F(n)=\sum_{d|n}\mu(n/d)G(d)\bigr)\,.
$$
Here, $\mu$ is the Möbius function: $\mu(m)=0$ unless $m$ is square-free, and in that case, $\mu(m)$ is $-1$ to a power equal to the number of prime factors of $m$. So $\mu(10)=1$, $\mu(11)=-1$, and $\mu(12)=0$. The proof may be tricky, but involves nothing fancy.
Now, you can see that the multiplicative version of the Inversion Formula is:
$$
G(n)=\prod_{d|n}F(d)\quad\Longrightarrow\quad F(n)=\prod_{d|n}G(d)^{\mu(n/d)}
$$
Well, my displayed formula is in exactly the format required by multiplicative Möbius, so that we have the resulting:
$$
\Phi(n)=\prod_{d|n}(x^d-1)^{\mu(n/d)}
$$
We apply this to see what $\Phi_9$, $\Phi_3$, and $\Phi_1$ are: it’s easier in this case, ’cause the only divisors of $9$ are $1,3,9$, and we get
$\Phi_1(x)=x-1$, $\Phi_3(x)=(x^3-1)(x-1)^{-1}=x^2+x+1$, and $\Phi_9(x)=(x^9-1)(x^3-1)^{-1}(x-1)^0=(x^9-1)/(x^3-1)=x^6+x^3+1$. And there are your three factors of $x^9-1$.
Well, justifying my first displayed formula is not hard, but involves using the group theory of cyclic groups, easy enough, but it’s late, I’m getting tired, and will skip that. But if you’re interested, let me know, and I’ll do an edit-addendum.
