Question

If $a=\sqrt[8]{6}-\sqrt[8]{5}, b=\sqrt[8]{6}+\sqrt[8]{5}, c=\sqrt[6]{6}+\sqrt[6]{5}, d=$ $\sqrt[4]{6}+\sqrt[4]{5}$, and $e=\sqrt{6}+\sqrt{5}$, then which of the following is a rational number?

• A. abcde
• B. $a b d e$
• C. $a b$
• D. $c d$

Answer 1

Answer: (B)

Given that, $a=\sqrt[8]{6}-\sqrt[8]{5}, b=\sqrt[8]{6}+\sqrt[8]{5}$
$c=\sqrt[6]{6}+\sqrt[6]{5} $
$d=\sqrt[4]{6}+\sqrt[4]{5}, e=\sqrt{6}+\sqrt{5}$
By inspection, we can reject $a b, c d$.
$a b=\sqrt[4]{6}+\sqrt[4]{5} $
$\therefore a b d=\sqrt{6}-\sqrt{5} \text { and } a b d e=6-5=1 .$
$\therefore$ abde is a rational number.