$n^4+n^2+1=(n^2-n+1)(n^2+n+1)=((n-1)^2+(n-1)+1)(n^2+n+1)$

A general form could be as follows $(a^2 + ab + b^2)(c^2 +cd + d^2) = (ac -bd)^2 + (ac - bd)(ad + bc + bd) + (ad + bc + bd)^2$.

EDIT ...

And the above is motivated by ...

AdBondEvent wrote: A general form could be as follows $(a^2 + ab + b^2)(c^2 +cd + d^2) = (ac -bd)^2 + (ac - bd)(ad + bc$$ \color{red}{+} $$bd) + (ad + bc $$\color{red}+$ $bd)^2$. just a typo.

$1^2+1+1=\frac{4^2+4+1}{2^2+2+1}$ $2^2+2+1=\frac{4^2+4+1}{1^2+1+1}\ (1)$ $3^2+3+1=\frac{9^2+9+1}{2^2+2+1} \ (2)$ By $(1)$ and $(2)$, we observe that $(n,a,b)=(2,2^2,2-1), (3,3^2,3-1)$ are solutions, so we conjecture $(n,a,b)=(n,n^2,n-1)$ is a general solution. And it is really true, because $((n-1)^2+(n-1)+1)(n^2+n+1)=(n^2-n+1)(n^2+n+1)=(n^2+1)^2-n^2=n^4+n^2+1=(n^2)^2+(n^2)+1$.

Solution: $(n^2+n+1)((n+1)^2+(n+1)+1)=(n^2+2n+1)^2+(n^2+2n+1)+1$. Generalisation: For any positive integer $n$, find positive integers $a,b$ such that $\frac{a^2+pa+q}{b^2+pb+q}=n^2+pn+q$. Here $p,q$ are fixed positive integers. Solution: $(n^2+pn+q)((n+1)^2+p(n+1)+q)=(n^2+(p+1)n+q)^2+p(n^2+(p+1)n+q)+q$.