Question

The greatest number among $\sqrt[3]{9},\sqrt[4]{11}$ and $\sqrt[6]{17}$ is

• A. $\sqrt[3]{9}$
• B. $\sqrt[4]{11}$
• C. $\sqrt[6]{17}$
• D. Cannot be determined.

Answer 1

The correct option is A

$\sqrt[3]{9}$

Explanation of the correct option.

Compute the required number:

Given: Three numbers $\sqrt[3]{9},\sqrt[4]{11}$ and $\sqrt[6]{17}$.

They can be written as $9^{\frac{1}{3}},{11}^{\frac{1}{4}}$ and ${17}^{\frac{1}{6}}$.

To compare irrational numbers, convert them to the same radical sign by taking L.C.M.

$9^{\frac{4}{12}},{11}^{\frac{3}{12}},{17}^{\frac{2}{12}}$

$$\begin{gathered} \Rightarrow {\left(9^2\right)}^{\frac{1}{12}},{\left({11}^3\right)}^{\frac{1}{12}},{\left({17}^2\right)}^{\frac{1}{12}} \\ \Rightarrow {\left(6561\right)}^{\frac{1}{12}},{\left(1331\right)}^{\frac{1}{12}},{\left(284\right)}^{\frac{1}{12}} \end{gathered}$$

Therefore, $\sqrt[3]{9}$ is the greatest number.

Hence, option $A$ is the correct answer.