If $\;\large\frac{1}{\sqrt{\alpha}}\;$ and $\;\large\frac{1}{\sqrt{\beta}}\;$ are the roots of the equation ,$\;ax^{2}+bx+c=0 \;(a \neq 0 , q ,b \in R)\;$, then the equation , $\;x(x+b^{3}) + (a^{3}-3abx)=0\;$ has roots :

$(a)\;\alpha^{\large\frac{3}{2}}\;and\; \beta^{\large\frac{3}{2}}\qquad(b)\;\alpha \beta^{\large\frac{3}{2}}\;and\; \beta \alpha^{\large\frac{3}{2}}\qquad(c)\;\sqrt{\alpha \beta} \;and\; \alpha \beta \qquad(d)\;\alpha^{-\large\frac{3}{2}}\;and\; \beta^{-\large\frac{3}{2}}$

The question is wrong, the correct equation is, ax^2+bx+1=0   , as the second equation involves no c.

Write the equation as x^2 + (b^3-3ab)x+a^3=0

The product of root of this equation is   [(âß)]^3/2   , which is only satiesfied by (a) optiion.