Let $y=\frac{x^8+x^4+1}{x^4+x^2+1}$. If $\frac{d y}{d x}=a x^3+b x$. Then,

$a=4, b=-2$

We have,

$y =\frac{x^8+x^4+1}{x^4+x^2+1}$

$\Rightarrow y =\frac{\left(x^4+1\right)^2-x^4}{x^4+x^2+1} = \frac{\left(x^4+1+x^2\right) \left(x^4+1-x^2\right)} {x^4+x^2+1}$

$\Rightarrow y =x^4-x^2+1$

$\Rightarrow \frac{d y}{d x}=4 x^3-2 x $

$\Rightarrow a x^3+b x=4 x^3-2 x$ for all x

$\Rightarrow a=4, b=-2$