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

Answer 1

The correct option is B $\frac{1}{4}$

$(\frac{{\alpha }^2}{\beta }{)}^{\frac{1}{3}}+(\frac{{\beta }^2}{\alpha }{)}^{\frac{1}{3}}$
$=\frac{{\alpha }^{\frac{2}{3}}}{{\beta }^{\frac{1}{3}}}+\frac{{\beta }^{\frac{2}{3}}}{{\alpha }^{\frac{1}{3}}}$
$=\frac{\alpha +\beta }{(\alpha \beta {)}^{\frac{1}{3}}}$...(i)
From the equation $8x^2-3x+27=0$
$\alpha +\beta =\frac{3}{8}$ and $\alpha \beta =\frac{27}{8}$
Substituting in the expression of i gives us
$=\frac{\alpha +\beta }{(\alpha \beta {)}^{\frac{1}{3}}}$
$=\frac{\frac{3}{8}}{(\frac{27}{8}{)}^{\frac{1}{3}}}$
$=\frac{\frac{3}{8}}{\frac{3}{2}}$
$=\frac{2}{8}$
$=\frac{1}{4}$.