If $\frac{1}{\sqrt{\alpha }}$ and $\frac{1}{\sqrt{\beta }}$ are the roots of the equation $a{x}^2+bx+1=0(a\ne 0,a,b,\in \mathbf{R}).$ then the equation,

$x\left(x+b^3\right)+\left(a^3-3abx\right)=0\text{ has roots: }$

 

The equation $x\left(x+b^3\right)+\left(a^3-3abx\right)=0$ can be 

written as

                      $x^2+\left(b^3-3ab\right)x+a^3=0$

We have

$\begin{array}{l}\frac{1}{\sqrt{\alpha }}+\frac{1}{\sqrt{\beta }}=-\frac{b}{a},\frac{1}{\sqrt{\alpha }\sqrt{\beta }}=\frac{1}{a}\\ \sqrt{\alpha }+\sqrt{\beta }=-b,\sqrt{\alpha }\sqrt{\beta }=a\\ \text{ Now }\\ \begin{array}{ll}-\left(b^3-3ab\right)& =(-b{)}^3-3a(-b)\\ & =(\sqrt{\alpha }+\sqrt{\beta }{)}^3-3(\sqrt{\alpha }+\sqrt{\beta })\sqrt{\alpha }\sqrt{\beta }\\ & =(\sqrt{\alpha }{)}^3+(\sqrt{\beta }{)}^3={\alpha }^{3/2}+{\beta }^{3/2}\end{array}\end{array}$

and $a^3={\alpha }^{3/2}{\beta }^{3/2}$ 

$\therefore$ The roots of the equation

$x^2+\left(b^2-3ab\right)x+a^3=0\text{ are }{\alpha }^{3/2},{\beta }^{3/2}$