Question

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

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

Answer 1

Answer: (B)

Since, $ \alpha ,\,\,\beta $ are the roots of the equation
$ 8{{x}^{2}}-3x+27=0 $
$ \therefore $ $ \alpha +\beta =\left( -\frac{3}{8} \right)=\frac{3}{8} $
$ \alpha \beta =\frac{27}{8}={{\left( \frac{3}{2} \right)}^{3}} $
Now, $ {{\left( \frac{{{\alpha }^{2}}}{\beta } \right)}^{1/3}}+{{\left( \frac{{{\beta }^{2}}}{\alpha } \right)}^{1/3}} $
$ =\frac{{{\alpha }^{2/3}}\cdot {{\alpha }^{1/3}}+{{\beta }^{2/3}}\cdot {{\beta }^{1/3}}}{{{(\alpha \beta )}^{1/3}}} $
$ =\frac{\alpha +\beta }{{{(\alpha \beta )}^{1/3}}} $
$ =\frac{\frac{3}{8}}{{{\left( \frac{27}{8} \right)}^{1/3}}}=\frac{3}{8}\cdot \frac{2}{3}=\frac{1}{4} $