Question : Which of the following given value is greater than $\sqrt[3]{12}$?
Option 1: $\sqrt[6]{121}$
Option 2: $\sqrt[12]{33214}$
Option 3: $\sqrt[5]{60}$
Option 4: $\sqrt[9]{1500}$
Correct Answer: $\sqrt[12]{33214}$
Solution :
Given number: $\sqrt[3]{12}$ $\sqrt[6]{121}$, $\sqrt[12]{33214}$ , $\sqrt[5]{60}$, $\sqrt[9]{1500}$
Now $\sqrt[6]{121}=\sqrt[6]{11^2}=\sqrt[3]{11}<\sqrt[3]{12}$
Also, $144>125$
⇒ $125 × 12^3<144 × 12^3$
⇒ $(5×12)^3<12^5$
⇒ $(60)^3<12^5$
⇒ $\sqrt[5]{60}<\sqrt[3]{12}$
Again, $1500<1728$
⇒ $(1500)^\frac{1}{9}<(1728)^\frac{1}{9}$
⇒ $\sqrt[9]{1500}<\sqrt[3]{12}$
Also, $33214>20736$
⇒ $(33214)^\frac{1}{12}>(20736)^\frac{1}{12}$
⇒ $\sqrt[12]{33214}>\sqrt[3]{12}$
Hence, the correct answer is $\sqrt[12]{33214}$.
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