If $\alpha$ and $\beta$ be two roots of the equation
x2 – 64x + 256 = 0. Then the value of
$${\left( {{{{\alpha ^3}} \over {{\beta ^5}}}} \right)^{1/8}} + {\left( {{{{\beta ^3}} \over {{\alpha ^5}}}} \right)^{1/8}}$$ is :
Solution
x<sup>2</sup> – 64x + 256 = 0
<br><br>$\alpha$ + $\beta$ = 64, $\alpha$$\beta$ = 256
<br><br>$${\left( {{{{\alpha ^3}} \over {{\beta ^5}}}} \right)^{1/8}} + {\left( {{{{\beta ^3}} \over {{\alpha ^5}}}} \right)^{1/8}}$$
<br><br>= $${{{\alpha ^{{3 \over 8}}}} \over {{\beta ^{{5 \over 8}}}}} + {{{\beta ^{{3 \over 8}}}} \over {{\alpha ^{{5 \over 8}}}}}$$
<br><br>= ${{\alpha + \beta } \over {{{\left( {\alpha \beta } \right)}^{{5 \over 8}}}}}$
<br><br>= ${{64} \over {{{\left( {256} \right)}^{{5 \over 8}}}}}$
<br><br>= 2
About this question
Subject: Mathematics · Chapter: Complex Numbers and Quadratic Equations · Topic: Complex Numbers and Argand Plane
This question is part of PrepWiser's free JEE Main question bank. 223 more solved questions on Complex Numbers and Quadratic Equations are available — start with the harder ones if your accuracy is >70%.