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

• A. 1
• B. $3 r$
• C. 4
• D. 2

Answer 1

Answer: (D)

$x^{2}-64 x+256=0$
$\alpha+\beta=64, \alpha \beta=256$
$\left(\frac{\alpha^{3}}{\beta^{5}}\right)^{1 / 8}+\left(\frac{\beta^{3}}{\alpha^{5}}\right)^{1 / 8}$
$=\frac{\alpha^{3 / 8}}{\beta^{5 / 8}}+\frac{\beta^{3 / 8}}{\alpha^{5 / 8}}$
$=\frac{\alpha+\beta}{(\alpha \beta)^{5 / 8}}$
$=\frac{64}{(256)^{5 / 8}}$
$=2$