Question

Arrange in descending order of magnitude $\sqrt[3]{32},\sqrt[6]{63},\sqrt[9]{94}$

• A. $\sqrt[3]{32}>\sqrt[9]{94}>\sqrt[6]{63}$
• B. $\sqrt[6]{63}>\sqrt[3]{32}>\sqrt[9]{94}$
• C. $\sqrt[9]{94}>\sqrt[6]{63}>\sqrt[3]{32}$
• D. None of these

Answer 1

The correct option is D None of these

Let us rewrite the given set of magnitudes $\sqrt[3]{32},\sqrt[6]{63},\sqrt[9]{94}$ as follows:

$\sqrt[3]{32}={\left(2\right)}^{\frac{1}{3}}\sqrt[6]{63}={\left(3\right)}^{\frac{1}{6}}\sqrt[9]{94}={\left(4\right)}^{\frac{1}{9}}$

We now take the LCM of the denominators of the powers to make the denominators same, then the above magnitudes will be:

$\sqrt[3]{32}={\left(2\right)}^{\frac{1\times 6}{3\times 6}}={\left(2\right)}^{\frac{6}{18}}={\left(2^6\right)}^{\frac{1}{18}}={\left(64\right)}^{\frac{1}{18}}=\sqrt[18]{1864}\sqrt[6]{63}={\left(3\right)}^{\frac{1\times 3}{6\times 3}}={\left(3\right)}^{\frac{3}{18}}={\left(3^3\right)}^{\frac{1}{18}}={\left(27\right)}^{\frac{1}{18}}=\sqrt[18]{1827}\sqrt[9]{94}={\left(4\right)}^{\frac{1\times 2}{9\times 2}}={\left(4\right)}^{\frac{2}{18}}={\left(4^2\right)}^{\frac{1}{18}}={\left(16\right)}^{\frac{1}{18}}=\sqrt[18]{1816}$

Therefore, the descending order is shown below:

$\sqrt[18]{1864}>\sqrt[18]{1827}>\sqrt[18]{1816}\Rightarrow \sqrt[3]{32}>\sqrt[6]{63}>\sqrt[9]{94}$

Hence, the descending order of the magnitude $\sqrt[3]{32},\sqrt[6]{63},\sqrt[9]{94}$ is $\sqrt[3]{32}>\sqrt[6]{63}>\sqrt[9]{94}$