3. Factorize: $a+(a+b) x+(a+2 b) x^{2}+(a+3 b) x^{3}+3 b x^{4}+2 b x^{5}+b x^{6}$.
3. Considering the fourth roots of unity other than 1, substituting $x=\varepsilon$ into the original expression yields $a+(a+b) \varepsilon+(a+2 b) \varepsilon^{2}+(a+3 b) \varepsilon^{3}+2 b \varepsilon^{5}+b \varepsilon^{6}=(a+3 b)\left(1+\varepsilon+\varepsilon^{2}+\varepsilon^{3}\right)=0$, we know that the original expression contains the factor $x^{3}+x^{2}+x+1$. Using polynomial division, we get the other factor $a+b x+b x^{2}+b x^{3}$. Therefore, the original expression $=(1+x)\left(1+x^{2}\right)\left(a+b x+b x^{2}+b x^{3}\right)$.