If two roots of the equation $(a-1)(x^2+x+1)^2-(a+1)(x^4+x^2+1)=0$ are real and distinct, then find the interval in which $a$ lies 
If two roots of the equation $(a-1)(x^2+x+1)^2-(a+1)(x^4+x^2+1)=0$ are real and distinct, then find the interval in which $a$ lies.    

My Approach:
I have expanded the equation to obtain a quartic equation which I am not able to factorize:
$$ x^4+(1-a)x^3+(2-a)x^2-ax+1=0 $$  
If only I could factorize it I would get two quadratic equations, one of which should have real roots. I know how to proceed further in order to find the interval in which a lies.   
But as for now I don't know how to proceed further. It would be great if I could get a hint to move forward.

The term $x^4 + x^2 + 1$ easily gets factorised as follows:
$$
x^4 + x^2 + 1 = (x^2 + x + 1)(x^2 - x + 1)
$$
Applying this in the original equation,
$$
(x^2 + x + 1)[(a - 1)(x^2 + x + 1) + (a + 1)(x^2 - x + 1)] = 0.
$$
Here $x^2 + x + 1 = 0$ is the quadratic with the complex roots. Focusing on the other quadratic, it can be simplified to:
$$
ax^2 - x + a = 0
$$
For this to have real and distinct roots, $1 - 4a^2 > 0 \implies a^2 < 1/4$ which simply gives us
$$
a \in \left (- \frac{1}{2}, \frac{1}{2}\right ).
$$
