Question

If $\alpha$ and $\beta$ be two roots of the equation $x^2-64x+256=0$
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. $3r$
• C. 4
• D. 2

Answer 1

Answer: (D)

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