Which one of the following set of surds is correct sequence of ascending order of their values?

\sqrt[4]{10}, \sqrt[3]{6}, \sqrt{3}

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\sqrt{3}, \sqrt[4]{10}, \sqrt[3]{6},

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\sqrt{3}, \sqrt{10}, \sqrt[3]{6},

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\sqrt[4]{10}, \sqrt{3}, \sqrt[3]{6}

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Solution

The correct option is B $\sqrt{3}, \sqrt[4]{10}, \sqrt[3]{6},$
$\sqrt[4]{10}, \sqrt[3]{6}, \sqrt{3}$
The order of the given irrational numbers are 2,3,4.
LCM of (2,3,4)=12
Now convert each irrational number as of order 12
$\sqrt[4]{10} = \sqrt[12]{10^{3}} = \sqrt[12]{1000}$
$\sqrt[3]{6} = \sqrt[12]{6^{4}} = \sqrt[12]{1296}$
$\sqrt{3} = \sqrt[12]{3^{6}} = \sqrt[12]{729}$
Hence, ascending order $\sqrt{3} < \sqrt[4]{10} < \sqrt[3]{6}$

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