If 1 α , 1 β are the roots of the equation a x ^ 2 + bx + 1 = 0, (a...

a{x}^{2} + bx + 1 = 0$has roots$\dfrac{1}{\sqrt{\alpha}}, \dfrac{1}{\sqrt{\beta}}$<br /><br />$⇒ \dfrac{1}{\sqrt{\alpha}} + \dfrac{1}{\sqrt{\beta}} = - \dfrac{b}{a} . . . \left(1\right)$<br /><br />$\dfrac{1}{\sqrt{\alpha}\sqrt{\beta}} = \dfrac{1}{a} . . . \left(2\right)$<br /><br />$x\left(x + {b}^{3}\right) + \left({a}^{3} - 3abx\right) = 0$<br /><br />$⇒ {x}^{2} + \left({b}^{3} - 3ab\right)x + {a}^{3} = 0$</p><p>Multiply and divide with${a}^{3}$, we get<br /><br />$⇒ {x}^{2} + {a}^{3}\left\{\left(\dfrac{{b}^{3}}{{a}^{3}}\right) - 3\dfrac{1}{a}\left(\dfrac{b}{a}\right)\right\}x + {a}^{3} = 0$<br />From equations \left(1\right) \text{ and } \left(2\right)$,
$⇒ {x}^{2} + {\left(\alpha\beta\right)}^{3 / 2}\left[- \dfrac{{\left({\alpha}^{1 / 2} + {\beta}^{1 / 2}\right)}^{3}}{{\left(\sqrt{\alpha\beta}\right)}^{3}} - 3\dfrac{1}{\sqrt{\alpha\beta}}\dfrac{\sqrt{\alpha} + \sqrt{\beta}}{\sqrt{\alpha\beta}}\right]x + {\left(\alpha\beta\right)}^{3 / 2} = 0$

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

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