Question

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

Answer 1

For $8x^2-3x+27=0$
$a+b=\frac{3}{8}$ and $ab=\frac{27}{8}$

Now ,$({\left(\frac{a^2}{b}\right)}^{\frac{1}{3}}+{\left(\frac{b^2}{a}\right)}^{\frac{1}{3}})={\left(\frac{a^3}{ab}\right)}^{\frac{1}{3}}+{\left(\frac{b^3}{ab}\right)}^{\frac{1}{3}}$
$=\frac{a+b}{(ab{)}^{\frac{1}{3}}}$
$=\frac{\frac{3}{8}}{{\left(\frac{27}{8}\right)}^{\frac{1}{3}}}$
$=\frac{1}{4}$