Question

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

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

Answer 1

The correct option is A

$\frac{1}{4}$

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

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

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