Which of the following is an irreducible factor of $x^{12}-1$ over $\mathbb Q$? Which of the following is an irreducible factor of $x^{12}-1$ over $\mathbb Q$?  


*

*$x^8+x^4+1$    

*$x^4+1$   

*$x^4-x^2+1$   

*$x^5-x^4+x^3-x^2+x-1$    


Is the answer $(1)$ correct? I am not sure which one is.
 A: By repeatedly using the simple factorizations:
$$x^{2n}-1=(x^n+1)(x^n-1)$$
and
$$x^{3n}-1=(x^{2n}+x^n+1)(x^n-1)$$
and
$$x^{3n}+1=(x^{2n}-x^n+1)(x^n+1)$$
We get:
$$x^{12}-1=(x^4-x^2+1)(x^2+1)(x^2-x+1)(x+1)(x^2+x+1)(x-1)$$
Ruling out every choice except for number 3:
$$x^4-x^2+1$$
Call this $p(x)$. There are many ways to show that $p(x)$ is irreducible. For example, we can note that it has no real roots, and hence it must be a product of two 2nd degree polynomials, if it is not irreducible. But $p(2)=p(-2)=13$ and $p(3)=p(-3)$ and $p(4)=p(-4)=241$, so it is a prime number for at least 6 different integer values of $x$. However, a product of two second degree polynomials with integer coefficients can only be prime for $4$ integer values of $x$, since one of the factors must be either $1$ or $-1$ for $p(x)$ to be prime. 
A: Warning: Probably too advanced for this question, but if you want to jump ahead.
If you know the cyclotomic polynomials, this factors into irreducibles as:
$$x^{12}-1=\Phi_1(x)\Phi_2(x)\Phi_3(x)\Phi_4(x)\Phi_6(x)\Phi_{12}(x)$$
Where $\Phi_d(x)$ is the polynomial having the primitive $d$th roots of unity as roots (the $d$th cyclotomic polynomial.)
This is because $x^{12}-1$ has every primitive $d$th root of unity as a root, for $d=1,2$.
So $$\begin{align}
\Phi_1(x)&=x-1\\
\Phi_2(x)&=\frac{x^2-1}{\Phi_1(x)}=x+1\\
\Phi_3(x)&=\frac{x^3-1}{\Phi_1(x)} = x^2+x+1\\
\Phi_4(x)&=\frac{x^4-1}{\Phi_2(x)\Phi_1(x)} = x^2+1\\
\Phi_6(x)&=\frac{x^6-1}{\Phi_3(x)\Phi_2(x)\Phi_1(x)} = x^2-x+1\\
\Phi_{12}(x)&=\frac{x^{12}-1}{\Phi_6(x)\Phi_4(x)\Phi_3(x)\Phi_2(x)\Phi_1(x)}
= x^4-x^2+1
\end{align}$$
Proving that Cyclotomic polynomials are irreducible is famously non-trivial.
A: Hint:
$$x^{12}-1 = (x^6-1)(x^6+1) = (x^3-1)(x^3+1)(x^6+1)=\cdots$$
Can you continue factorizing? (answer: yes; you can factorize $(x^6+1)$ using complex numbers... but the factors maybe $\not\in\Bbb Q[x]$)
