If the two roots of the equation, (a)(x⁴+x²+1)+(a+1)(x²+x+1)²=0 are real and distinct, then the set of all values of a is:

Given equation is (a)(x⁴+x²+1)+(a+1)(x²+x+1)²=0
→(a)(x⁴+2x²+1−x²)+(a+1)(x²+x+1)²=0
→(a)(x²+x+1)(x²−x+1)+(a+1)(x²+x+1)²=0
→(x²+x+1)[(a)(x²−x+1)+(a+1)(x²+x+1)]=0
→(x²+x+1)[ax²+x+a]=0
Clearly x²+x+1=0 won't have real solution.
So for given equation to posses two real and distinct solution discriminant of ax²+x+a=0 has to be >0
→(1)²−4a²>0
→a²<1/4
→(−1/2,0)∪(0,1/2)