Cyclotomic polynomials

Roots. If $\omega\neq1$ and $\omega^3=1$, then $(\omega-1)(\omega^2+\omega+1)=0$, so $\omega$ is a root of $X^2+X+1$.

Discriminant. $\Delta=1-4=-3<0$, so the roots are complex: $\omega=(-1\pm i\sqrt3)/2$, with $|\omega|=1$ and $\omega^3=1$.

Irreducibility over $\mathbb{Q}$. A quadratic with no rational root is irreducible over $\mathbb{Q}$. $X^2+X+1$ has no rational root (candidates $\pm1$ give $3$ and $1$).

Conclusion. $\Phi_3(X)=X^2+X+1$ is irreducible over $\mathbb{Q}$, with roots the primitive cube roots of unity.