If the two roots of the equation, a - 1   x 4 + x 2 + 1 + a + 1 x 2 + x + 1 2 = 0 are real and distinct…

$\left(a-1\right)\left(x^4+x^2+1\right)+\left(a+1\right){\left(x^2+x+1\right)}^2=0$

$\left(x^2+x+1\right) \left[\left(a-1\right)\left(x^2-x+1\right)+\left(a+1\right)(x^2+ x+1)\right]=0$

$\left(x^2+x+1\right)\left[2a{x}^2+2x+2a\right]=0$

$\Rightarrow \left(x^2+x+1\right) \left[a{x}^2+x+a\right]=0$

If two Roots are real

Then, roots of $a{x}^2+x+a=0$ should be real & distinct.

$\therefore 1-4a^2>0$

$\left(1-2a\right)\left(1+2a\right)>0$

As $a\ne 0$ therefore

$a\in \left(-\frac{1}{2},0\right) \cup \left(0,\frac{1}{2}\right)$