Ta có: \(x^4+ax^2+b\)
\(=\left(x^4-x^3+x^2\right)+\left(x^3-x^2+x\right)+\left(ax^2-ax+a\right)+b+ax-a-x\)
\(=x^2\left(x^2-x+1\right)+x\left(x^2-x+1\right)+a\left(x^2-x+1\right)+\left(a-1\right)x+\left(b-a\right)\)
\(=\left(x^2+x+a\right)\left(x^2-x+1\right)+\left(a-1\right)x+\left(b-a\right)\)
Vì \(\left(x^2+x+a\right)\left(x^2-x+1\right)⋮\left(x^2-x+1\right)\)
nên để \(\left(x^2+x+a\right)\left(x^2-x+1\right)+\left(a-1\right)x+\left(b-a\right)⋮\left(x^2-x+1\right)\)
thì: \(\left\{{}\begin{matrix}a-1=0\\b-a=0\end{matrix}\right.\Rightarrow a=b=1\)
\(x^4+ax^2+b⋮x^2-x+1\)
=>\(x^4-x^3+x^2+x^3-x^2+x+ax^2-x+b⋮x^2-x+1\)
=>\(\left\{{}\begin{matrix}a=1\\b=1\end{matrix}\right.\)