Question

Compare the following pair of surds:

$\sqrt[3]{36},\sqrt[4]{48}$

• A. $\sqrt[3]{36}>\sqrt[4]{48}$
• B. $\sqrt[3]{36}>\sqrt[4]{44}$
• C. $\sqrt[3]{36}>\sqrt[3]{38}$
• D. $\sqrt[3]{34}>\sqrt[4]{48}$

Answer 1

Answer: (A)

$\sqrt[3]{36}>\sqrt[4]{48}$

$\sqrt[3]{36},\sqrt[4]{48}$
To compare, the order of the surds must be same.
$=>6^{\frac{1}{3}},8^{\frac{1}{4}}$
$=>6^{\frac{4}{12}},8^{\frac{3}{12}}$
$=>\sqrt[12]{126^4},\sqrt[12]{128^3}$
$=>\sqrt[12]{121296},\sqrt[12]{12512}$
Therefore,
$=>1296>512$
$=>\sqrt[12]{126^4}>\sqrt[12]{128^3}$
$=>\sqrt[3]{36}>\sqrt[4]{48}$