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)

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