Mathematics · 2020
JEE Main · 6 September 2020, Shift 1 · Q52
If α and β be two roots of the equation x^2-64 x+256=0. Then the value of ((α^3)/(β^5))^1 / 8+((β^3)/(α^5))^1 / 8 is:
If $\displaystyle \alpha$ and $\displaystyle \beta$ be two roots of the equation $\displaystyle x^2-64 x+256=0$. Then the value of $\displaystyle \left(\frac{\alpha^3}{\beta^5}\right)^{1 / 8}+\left(\frac{\beta^3}{\alpha^5}\right)^{1 / 8}$ is :
Official answer
(3)
$\displaystyle 2$
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JEE Main 2020 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.