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

Answer 1

The correct option is C

$\frac{1}{4}$

Given: $\alpha ,\beta$ are the roots of the equation $8x^2-3x+27=0$,
Using the relation between roots and coefficients, we get
$\alpha +\beta =\frac{3}{8},\alpha \beta =\frac{27}{8}\therefore {\left(\frac{{\alpha }^2}{\beta }\right)}^{\frac{1}{3}}+{\left(\frac{{\beta }^2}{\alpha }\right)}^{\frac{1}{3}}=\frac{({\alpha }^3{)}^{\frac{1}{3}}+({\beta }^3{)}^{\frac{1}{3}}}{(\alpha \beta {)}^{\frac{1}{3}}}=\frac{\alpha +\beta }{(\alpha \beta {)}^{\frac{1}{3}}}=\frac{3/8}{(27/8{)}^{\frac{1}{3}}}=\frac{3/8}{3/2}=\frac{1}{4}$.