If  $\alpha ,\beta$ are the roots of the equation $8x^2-3x+27=0$Then the value of ${\left(\frac{\alpha }{{\beta }^2}\right)}^{\frac{1}{3}}+{\left(\frac{\beta }{{\alpha }^2}\right)}^{\frac{1}{3}} is$

If  $\alpha , \beta$are the roots of the equation $8x^2-3x+27=0$ then

$\alpha +\beta =\frac{3}{8} , \alpha \beta =\frac{27}{8}$

now $$\begin{gathered} {\left(\frac{\alpha }{{\beta }^2}\right)}^{\frac{1}{3}}+{\left(\frac{\beta }{{\alpha }^2}\right)}^{\frac{1}{3}} =\frac{{\alpha }^{{\displaystyle \frac{1}{3}}}}{{\beta }^{{{\displaystyle \frac{2}{3}}}^{ }}}+\frac{{\beta }^{{\displaystyle \frac{1}{3}}}}{{\alpha }^{{\displaystyle \frac{2}{3}}}} \\ =\frac{{\alpha }^{{\displaystyle \frac{1}{3}}}.{\alpha }^{{\displaystyle \frac{2}{3}}}+{\beta }^{{\displaystyle \frac{1}{3}}}{\beta }^{{\displaystyle \frac{2}{3}}}}{{\alpha }^{{\displaystyle \frac{2}{3}}}{\beta }^{{\displaystyle \frac{2}{3}}}} \\ =\frac{\alpha +\beta }{{\left(\alpha \beta \right)}^{{\displaystyle \frac{2}{3}}}} \\ =\frac{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$8$}\right.}}{{\left({\displaystyle \frac{27}{8}}\right)}^{{\displaystyle \frac{2}{3}}}}=\frac{1}{6} \end{gathered}$$