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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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