Question

Arrange the following in ascending order of magnitude:

$\sqrt[3]{32},\sqrt[4]{43},\sqrt[3]{34}$

• A. $\sqrt[3]{32}>\sqrt[4]{43}>\sqrt[3]{34}$
• B. $\sqrt[3]{32}>\sqrt[4]{43}<\sqrt[3]{34}$
• C. $\sqrt[3]{32}<\sqrt[4]{43}>\sqrt[3]{34}$
• D. $\sqrt[3]{32}<\sqrt[4]{43}<\sqrt[3]{34}$

Answer 1

The correct option is D $\sqrt[3]{32}<\sqrt[4]{43}<\sqrt[3]{34}$

$\sqrt[3]{32},\sqrt[4]{43},\sqrt[3]{34}$
To compare, the order of the surds must be same.
$=>2^{\frac{1}{3}},3^{\frac{1}{4}},4^{\frac{1}{3}}$
$=>2^{\frac{4}{12}},3^{\frac{3}{12}},4^{\frac{4}{12}}$
$=>\sqrt[12]{122^4},\sqrt[12]{123^3},\sqrt[12]{124^4}$
$=>\sqrt[12]{1216},\sqrt[12]{1227},\sqrt[12]{12256}$
Therefore,
$=>16<27<256$
$=>\sqrt[12]{122^4}<\sqrt[12]{123^3}<\sqrt[12]{124^4}$
$=>\sqrt[3]{32}<\sqrt[4]{43}<\sqrt[3]{34}$