If α,ß are the roots of the equation 8x²–3x+27=0, then the value of ${\left(\frac{\mathrm{\alpha }}{{\mathrm{\beta }}^2}\right)}^{\frac{1}{3}}+{\left(\frac{\mathrm{\beta }}{{\mathrm{\alpha }}^2}\right)}^{\frac{1}{3}}$ is

For the given equation $8x^2-3x+27=0$

If $\alpha ,\beta$  are the roots of the equation 

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

Hence,  ${\left(\frac{\mathrm{\alpha }}{{\mathrm{\beta }}^2}\right)}^{\frac{1}{3}}+{\left(\frac{\mathrm{\beta }}{{\mathrm{\alpha }}^2}\right)}^{\frac{1}{3}}=\frac{\alpha +\beta }{{\left(\alpha \beta \right)}^{{\displaystyle \frac{2}{3}}}}$
$\frac{\mathrm{\alpha }+\mathrm{\beta }}{(\mathrm{\alpha \beta }{)}^{2/3}}=\frac{\frac{3}{8}}{{\left(\frac{27}{8}\right)}^{2/3}}=\frac{1}{6}$