ax2+bx+1=0 has roots α1,β1
⇒α1+β1=−ab...(1)
αβ1=a1...(2)
x(x+b3)+(a3−3abx)=0
⇒x2+(b3−3ab)x+a3=0
Multiply and divide with a3, we get
⇒x2+a3(a3b3)−3a1(ab)x+a3=0
From equations (1)&(2),
⇒x2+(αβ)3/2[−(αβ)3(α1/2+β1/2)3−3αβ1αβα+β]x+(αβ)3/2=0
⇒x2−(α3/2+β3/2)x+α3/2β3/2=0
Roots are α23 and β23.