x^4+ax^2+b chia hết cho x^2+x+1
Đặt $x^4+ax^2+b=(x^2+x+1)(x^2+cx+d)$
$\Leftrightarrow x^4+ax^2+b=x^4+(c+1)x^3+(c+d+1)x^2+(c+d)x+d$
$\Leftrightarrow \left\{\begin{matrix} c+1=0 \\ c+d+1=a \\ c+d=0 \\ d=b \end{matrix} \right.
\Leftrightarrow \left\{\begin{matrix} a=b=d=1 \\ c=-1 \end{matrix} \right.$