If the roots of equation $1(\mathrm{a}-1){\left({\mathrm{x}}^2+\mathrm{x}+1\right)}^2=(\mathrm{a}+1)\left({\mathrm{x}}^4+{\mathrm{x}}^2+1\right)$ are real and distinct, then the value of $a \in$

$\begin{array}{l}{\mathrm{x}}^4+{\mathrm{x}}^2+1={\left({\mathrm{x}}^2+1\right)}^2-{\mathrm{x}}^2\\ =\left({\mathrm{x}}^2+\mathrm{x}+1\right)\left({\mathrm{x}}^2-\mathrm{x}+1\right)\\ {\mathrm{x}}^2+\mathrm{x}+1={\left(\mathrm{x}+\frac{1}{2}\right)}^2+\frac{3}{4}\ne 0\mathrm{\forall }\mathrm{x}\end{array}$

Therefore, we can cancel this factor and we get

$\begin{array}{l}(\mathrm{a}-1)\left({\mathrm{x}}^2-\mathrm{x}+1\right)=(\mathrm{a}+1)\left({\mathrm{x}}^2-\mathrm{x}+1\right)\\ \mathrm{or} {\mathrm{x}}^2-\mathrm{ax}+1=0\end{array}$

It has real and distinct roots if $\mathrm{D}={\mathrm{a}}^2-4>0$