Question

the greatest among ³√9,⁴√11,and 6root of17 is
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Answer 1

Answer:

$\sqrt[3]{9}$ is greater

Step-by-step explanation:

$\sqrt[3]{9},\sqrt[4]{11},\sqrt[6]{17}$

L.C.M(3,4,6)=12

Now,

$\sqrt[3]{9}=\sqrt[12]{9^4}=\sqrt[12]{6561}$

$\sqrt[4]{11}=\sqrt[12]{11^3}=\sqrt[12]{1331}$

$\sqrt[6]{17}=\sqrt[12]{17^2}=\sqrt[12]{289}$

Now,

$289<1331<6561$

$\implies\:\sqrt[12]{289}<\sqrt[12]{1331}<\sqrt[12]{6561}$

$\implies\:\sqrt[6]{17}<\sqrt[4]{11}<\sqrt[3]{9}$

$\therefore\:\sqrt[3]{9}$ is greater.