The greatest number among sqrt[3]{9}, sqrt[4] | vedclass

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Question

(A) To compare the numbers $\sqrt[3]{9}, \sqrt[4]{11}, 	ext{ and } \sqrt[6]{17}$,we express them with a common index.The indices are $3, 4, 	ext{ an

Vedclass · Accepted Answer

$\sqrt[3]{9}$

Vedclass · Answer

(A) To compare the numbers $\sqrt[3]{9}, \sqrt[4]{11}, 	ext{ and } \sqrt[6]{17}$,we express them with a common index.The indices are $3, 4, 	ext{ and } 6$. The $L.C.M.$ of $3, 4, 6$ is $12$.Converting each to the $12^{th}$ root:$\sqrt[3]{9} = 9^{1/3} = (9^4)^{1/12} = (6561)^{1/12}$$\sqrt[4]{11} = 11^{1/4} = (11^3)^{1/12} = (1331)^{1/12}$$\sqrt[6]{17} = 17^{1/6} = (17^2)^{1/12} = (289)^{1/12}$Comparing the values inside the $12^{th}$ root: $6561 > 1331 > 289$.Therefore,$\sqrt[3]{9}$ is the greatest number.

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

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