If 1 α , 1 β are the roots of the equation a x 2 + b x + 1 = 0 ,   a ≠ 0 ,   a , b…

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

$\Rightarrow \frac{1}{\sqrt{\alpha }}+\frac{1}{\sqrt{\beta }}=-\frac{b}{a} ...\left(1\right)$

$\frac{1}{\sqrt{\alpha }\sqrt{\beta }}=\frac{1}{a} ...\left(2\right)$

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

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

Multiply and divide with $a^3$, we get

$\Rightarrow x^2+a^3\left\{\left(\frac{b^3}{a^3}\right)-3\frac{1}{a}\left(\frac{b}{a}\right)\right\}x+a^3=0$
From equations $\left(1\right) \& \left(2\right)$,
$\Rightarrow x^2+{\left(\alpha \beta \right)}^{3/2}\left[-\frac{{\left({\alpha }^{1/2}+{\beta }^{1/2}\right)}^3}{{\left(\sqrt{\alpha \beta }\right)}^3}-3\frac{1}{\sqrt{\alpha \beta }}\frac{\sqrt{\alpha }+\sqrt{\beta }}{\sqrt{\alpha \beta }}\right]x+{\left(\alpha \beta \right)}^{3/2}=0$

$\Rightarrow x^2-\left({\alpha }^{3/2}+{\beta }^{3/2}\right)x+{\alpha }^{3/2}{\beta }^{3/2}=0$

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