Solve this following

Question:

If $\alpha$ and $\beta$ be two roots of the equation

$x^{2}-64 x+256=0$

Then the value of $\left(\frac{\alpha^{3}}{\beta^{5}}\right)^{\frac{1}{8}}+\left(\frac{\beta^{3}}{\alpha^{5}}\right)^{\frac{1}{8}}$ is

  1. 1

  2. 3

  3. 4

  4. 2


Correct Option: , 4

Solution:

$x^{2}-64 x+256=0$

$\alpha+\beta=64, \alpha \beta=256$

$\left(\frac{\alpha^{3}}{\beta^{5}}\right)^{1 / 8}+\left(\frac{\beta^{3}}{\alpha^{5}}\right)^{1 / 8}$

$=\frac{\alpha^{3 / 8}}{\beta^{5 / 8}}+\frac{\beta^{3 / 8}}{\alpha^{5 / 8}}$

$=\frac{\alpha+\beta}{(\alpha \beta)^{5 / 8}}$

$=\frac{64}{(256)^{5 / 8}}$

$=2$

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