If 1 √α , 1 √β are the roots of the equation ax^2+bx+1=0,(a≠ 0,a,b∈ R), then the equation x(x+b^3)+(a^3-3abx)=0 has roots:

$a{x}^{2}+bx+1=0$ax2+bx+1=0 has roots $\frac{1}{\sqrt{\alpha }},\frac{1}{\sqrt{\beta }}$α​1​,β​1​

$$\Rightarrow \frac{1}{\sqrt{\alpha }}+\frac{1}{\sqrt{\beta }}=-\frac{b}{a}...(1)$$⇒α​1​+β​1​=−ab​...(1)

$$\frac{1}{\sqrt{\alpha }\sqrt{\beta }}=\frac{1}{a}...(2)$$α​β​1​=a1​...(2)

$$x(x+{b}^{3})+({a}^{3}-3abx)=0$$x(x+b3)+(a3−3abx)=0

$$\Rightarrow {x}^{2}+({b}^{3}-3ab)x+{a}^{3}=0$$⇒x2+(b3−3ab)x+a3=0

Multiply and divide with ${a}^{3}$a3, we get

$$\Rightarrow {x}^{2}+{a}^{3}{(\frac{{b}^{3}}{{a}^{3}})-3\frac{1}{a}(\frac{b}{a})}x+{a}^{3}=0$$⇒x2+a3(a3b3​)−3a1​(ab​)x+a3=0

From equations (1)&(2),

$$\Rightarrow {x}^{2}+{(\alpha \beta )}^{3/2}[-\frac{{({\alpha }^{1/2}+{\beta }^{1/2})}^{3}}{{(\sqrt{\alpha \beta })}^{3}}-3\frac{1}{\sqrt{\alpha \beta }}\frac{\sqrt{\alpha }+\sqrt{\beta }}{\sqrt{\alpha \beta }}]x+{(\alpha \beta )}^{3/2}=0$$⇒x2+(αβ)3/2[−(αβ​)3(α1/2+β1/2)3​−3αβ​1​αβ​α​+β​​]x+(αβ)3/2=0

$$\Rightarrow {x}^{2}-({\alpha }^{3/2}+{\beta }^{3/2})x+{\alpha }^{3/2}{\beta }^{3/2}=0$$⇒x2−(α3/2+β3/2)x+α3/2β3/2=0

Roots are ${\alpha }^{\frac{3}{2}}\text{ and }{\beta }^{\frac{3}{2}}$α23​ and β23​.