Question : If the numbers $\sqrt[3]{9}, \sqrt[4]{20}$ and $\sqrt[6]{25}$ are arranged in ascending order, then the right arrangement is:
Option 1: $\sqrt[6]{25}<\sqrt[4]{20}< \sqrt[3]{9}$
Option 2: $\sqrt[3]{9}<\sqrt[4]{20}<\sqrt[6]{25}$
Option 3: $\sqrt[4]{20}<\sqrt[6]{25}<\sqrt[3]{9}$
Option 4: $\sqrt[6]{25}<\sqrt[3]{9}<\sqrt[4]{20}$
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Correct Answer: $\sqrt[6]{25}<\sqrt[3]{9}<\sqrt[4]{20}$
Solution :
Given: The numbers are $\sqrt[3]{9}$, $\sqrt[4]{20}$, $\sqrt[6]{25}$.
LCM of 3, 4, and 6 = 12
$⇒ \sqrt[3]{9}=9^{\frac{1}{3}}=9^{\frac{4}{12}}=\sqrt[12]{6561}$
$⇒ \sqrt[4]{20}=20^{\frac{1}{4}}=20^{\frac{3}{12}}=\sqrt[12]{8000}$
$⇒ \sqrt[6]{25}=25^{\frac{1}{6}}=25^{\frac{2}{12}}=\sqrt[12]{625}$
By comparing these values, we get:
$\sqrt[12]{625}<\sqrt[12]{6561}<\sqrt[12]{8000}$
$⇒\sqrt[6]{25}<\sqrt[3]{9}<\sqrt[4]{20}$
Hence, the correct answer is $\sqrt[6]{25}<\sqrt[3]{9}<\sqrt[4]{20}$.
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