Question

$\alpha ,\beta$ are the roots of the equation $8x^2-3x+27=0$ , find the value of $(\frac{\alpha }{{\beta }^2}{)}^{\frac{1}{3}}+(\frac{\beta }{{\alpha }^2}{)}^{\frac{1}{3}}$

Answer 1

Given that,

$8x^2-3x+27=0$

$a=8$

$b=-3$

$c=27$

$\alpha +\beta =\frac{-b}{a}$

$=\frac{3}{8}$

$\alpha .\beta =\frac{c}{a}=\frac{27}{8}$

Now, ${\left(\frac{\alpha }{{\beta }^2}\right)}^{\frac{1}{3}}+{\left(\frac{\beta }{{\alpha }^2}\right)}^{\frac{1}{3}}$

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

$=\frac{\frac{3}{8}}{{\left(\frac{27}{8}\right)}^{\frac{2}{5}}}$

$=\frac{\frac{3}{8}}{\left|{\left(\frac{3}{2}\right)}^3\right|}$

$=\frac{\frac{3}{8}}{\frac{9}{4}}=\frac{3}{8}\times \frac{4}{9}$

Then value of ${\left(\frac{\alpha }{{\beta }^2}\right)}^{\frac{1}{3}}+{\left(\frac{\beta }{{\alpha }^2}\right)}^{\frac{1}{3}}=\frac{1}{6}$.