Question

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

• A. $\frac{1}{3}$
• B. $\frac{1}{4}$
• C. $\frac{1}{5}$
• D. $\frac{1}{6}$

Answer 1

Answer: (B)

$\alpha +\beta =\frac{3}{8},\alpha \beta =\frac{27}{8}$
${\left(\frac{{\alpha }^2}{\beta }\right)}^{\frac{1}{3}}+{\left(\frac{{\beta }^2}{\alpha }\right)}^{\frac{1}{3}}=\frac{{\left({\alpha }^3\right)}^{\frac{1}{3}}+{\left({\beta }^3\right)}^{\frac{1}{3}}}{(\alpha \beta {)}^{\frac{1}{3}}}$
$\frac{\alpha +\beta }{(\alpha \beta {)}^{\frac{1}{3}}}=\frac{\frac{3}{8}}{{\left(\frac{27}{8}\right)}^{\frac{1}{3}}}=\frac{\frac{3}{8}}{\frac{3}{2}}=\frac{1}{4}$