Question

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

• A. $\frac{1}{3}$
• B. $\frac{1}{4}$
• C. $\frac{7}{2}$
• D. $4$

Answer 1

Answer: (B)

Since, $\alpha ,\beta$ are the roots of the equation
$8x^2-3x+27=0$
$\therefore$ $\alpha +\beta =\left(-\frac{3}{8}\right)=\frac{3}{8}$
$\alpha \beta =\frac{27}{8}={\left(\frac{3}{2}\right)}^3$
Now, ${\left(\frac{{\alpha }^2}{\beta }\right)}^{1/3}+{\left(\frac{{\beta }^2}{\alpha }\right)}^{1/3}$
$=\frac{{\alpha }^{2/3}\cdot {\alpha }^{1/3}+{\beta }^{2/3}\cdot {\beta }^{1/3}}{{(\alpha \beta )}^{1/3}}$
$=\frac{\alpha +\beta }{{(\alpha \beta )}^{1/3}}$
$=\frac{\frac{3}{8}}{{\left(\frac{27}{8}\right)}^{1/3}}=\frac{3}{8}\cdot \frac{2}{3}=\frac{1}{4}$