Question

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

Solution

## Given that,$$8{x^2} - 3x + 27 = 0$$$$a=8$$$$b=-3$$$$c=27$$$$\alpha + \beta = \dfrac{{ - b}}{a}$$$$= \dfrac{3}{8}$$$$\alpha .\beta = \dfrac{c}{a} = \dfrac{{27}}{8}$$Now, $${\left( {\dfrac{\alpha }{{{\beta ^2}}}} \right)^{\dfrac{1}{3}}} + {\left( {\dfrac{\beta }{{{\alpha ^2}}}} \right)^{\dfrac{1}{3}}}$$$$= \dfrac{{\alpha + \beta }}{{{{(\alpha \beta )}^{\dfrac{2}{3}}}}}$$$$= \dfrac{{\dfrac{3}{8}}}{{{{\left( {\dfrac{{27}}{8}} \right)}^{\dfrac{2}{5}}}}}$$$$= \dfrac{{\dfrac{3}{8}}}{{\left| {{{\left( {\dfrac{3}{2}} \right)}^3}} \right|}}$$$$= \dfrac{{\dfrac{3}{8}}}{{\dfrac{9}{4}}} = \dfrac{3}{8} \times \dfrac{4}{9}$$Then value of $${\left( {\dfrac{\alpha }{{{\beta ^2}}}} \right)^{\dfrac{1}{3}}} + {\left( {\dfrac{\beta }{{{\alpha ^2}}}} \right)^{\dfrac{1}{3}}} = \dfrac{1}{6}$$.Mathematics

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