(a−1)(x4+x2+1)+(a+1)(x2+x+1)2=0
(x2+x+1)[(a−1)(x2−x+1)+(a+1)(x2+x+1)]=0
(x2+x+1)[2ax2+2x+2a]=0
⇒(x2+x+1)[ax2+x+a]=0
If two Roots are real
Then, roots of ax2+x+a=0 should be real & distinct.
∴1−4a2>0
(1−2a)(1+2a)>0

As a=0 therefore
a∈(−21,0)∪(0,21)