Irreducibility of $x^{11}+x^{10}+x^{9}+\dots+x+1$ 
Prove that$$g(x)=x^{11}+x^{10}+x^{9}+\cdots+x+1$$ is irreducible over $\mathbb{Q}$.

This is my (incorrect-see comments) attempt
$$g(x)=\frac{(x-1)(x^{11}+x^{10}+x^{9}+\cdots+x+1)}{x-1}=\frac{x^{12}-1}{x-1},$$
So$$g(x+1)=\frac{(x+1)^{12}-1}{x}=x^{11}+12x^{10}+66x^{9}+\cdots+12.$$ By Eisenstein criteria this polynomial is irreducible over $\mathbb{Q}[x]$ since $3$ divides $a_{i}$ where $i=0,1,2...10$, but $3^{2}$ does not divide $a_{0}$ and $3$ does not divide $a_{11}$.
Assume for contradiction $g(x)$ is reducible which means $\exists p(x),q(x)\in\mathbb{Q}$ both non-unit such that $g(x)=p(x)q(x)$. Hence $$g(x+1)=p(x+1)q(x+1)\Longrightarrow g(x+1)$$is reducible. A contradiction.
I am curious to know how to prove this result.

Hint:
It is not true because:
$$
(x^{12}-1)=(x^6+1)(x^6-1)=(x^6+1)(x^3+1)(x^3-1)=
$$
$$
=(x^6+1)(x+1)(x^2-x+1)(x-1)(x^2+x+1)
$$
and
$$
(x^{12}-1)=(x-1)(x^{11}+x^{10}+ \cdots +x+1)
$$
