Which of the following given value is greater than $$\root 3 \of {12} \,?$$

Solution (By Examveda Team)

$$\eqalign{ & {\text{Given, }}\root 3 \of {12} \cr & {\bf{Concept \,used:}} \cr & {\text{If }}a > b,{\text{ then:}} \cr & {a^{\frac{1}{3}}} > {b^{\frac{1}{3}}} \cr & {\bf{Calculation:}} \cr & {\text{Option}}\left( {\text{A}} \right):\root 9 \of {1500} = {\left( {1500} \right)^{\frac{1}{9}}} \cr & \Rightarrow \root 3 \of {12} = {12^{\frac{1}{3}}} \cr & \Rightarrow {\text{LCM}}\left( {9,\,3} \right) = 9 \cr & \Rightarrow {\left( {12} \right)^{\frac{1}{3} \times 9}}{\text{ and }}{\left( {1500} \right)^{\frac{1}{9} \times 9}} \cr & i.e.,{\left( {12} \right)^3}{\text{ and }}{\left( {1500} \right)^1} \cr & i.e.,\left( {1728} \right){\text{ and }}\left( {1500} \right) \cr & {\text{Since, }}1728 > 1500 \cr & \Rightarrow {\left( {12} \right)^{\frac{1}{3}}} > {\left( {1500} \right)^{\frac{1}{9}}} \cr & {\text{Option}}\left( {\text{B}} \right):\root 5 \of {60} = {\left( {60} \right)^{\frac{1}{5}}} \cr & \Rightarrow \root 3 \of {12} = {12^{\frac{1}{3}}} \cr & \Rightarrow {\text{LCM}}\left( {5,\,3} \right) = 15 \cr & \Rightarrow {\left( {12} \right)^{\frac{1}{3} \times 15}}{\text{ and }}{\left( {60} \right)^{\frac{1}{5} \times 15}} \cr & i.e.,{\left( {12} \right)^5}{\text{ and }}{\left( {60} \right)^3} \cr & {12^2} \times {12^3}{\text{ and }}{12^3} \times {5^3} \cr & 144 \times {12^3}{\text{ and }}125 \times {12^3} \cr & {\text{1}}{{\text{2}}^3}{\text{ is common both but }}114 > 125 \cr & \Rightarrow {\text{Hence }}{12^5} > {60^3}{\text{ or }}{12^{\frac{1}{3}}} > {60^{\frac{1}{5}}} \cr & {\text{Option}}\left( {\text{C}} \right):\root 6 \of {121} = {\left( {121} \right)^{\frac{1}{6}}} \cr & \Rightarrow \root 3 \of {12} = {12^{\frac{1}{3}}} \cr & \Rightarrow {\text{LCM}}\left( {6,\,3} \right) = 6 \cr & \Rightarrow {\left( {12} \right)^{\frac{1}{3} \times 6}}{\text{ and }}{\left( {121} \right)^{\frac{1}{6} \times 6}} \cr & i.e.,{\left( {12} \right)^2}{\text{ and }}{\left( {121} \right)^1} \cr & i.e.,{\left( {12} \right)^2}{\text{ and }}{\left( {11} \right)^2} \cr & {\text{Since, }}12 > 11 \cr & \Rightarrow {\left( {12} \right)^{\frac{1}{3}}} > {\left( {121} \right)^{\frac{1}{6}}} \cr & {\text{Option}}\left( {\text{D}} \right):\root {12} \of {33214} = {\left( {33214} \right)^{\frac{1}{{12}}}} \cr & \Rightarrow \root 3 \of {12} = {12^{\frac{1}{3}}} \cr & \Rightarrow {\text{LCM}}\left( {12,\,3} \right) = 12 \cr & \Rightarrow {\left( {12} \right)^{\frac{1}{3} \times 12}}{\text{ and }}{\left( {33214} \right)^{\frac{1}{{12}} \times 12}} \cr & i.e.,{\left( {12} \right)^4}{\text{ and }}{\left( {33214} \right)^1} \cr & i.e.,{\left( {20736} \right)^1}{\text{ and }}{\left( {33214} \right)^1} \cr & {\text{Since, }}33214 > 20736 \cr & \Rightarrow {\left( {12} \right)^{\frac{1}{3}}} > {\left( {33214} \right)^{\frac{1}{{12}}}} \cr & \therefore {\text{ The required greatest value is }}\root {12} \of {33214} . \cr} $$