[\sqrt[3]{3}] Which is larget √2 or \sqrt[3]{3}?

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Aptitude

TNPSC G4

10th CBSE Maths

Which is larget √2 or $\sqrt[3]{3}$?

√2

$\sqrt[3]{3}$

Both are equal

None of these

Explanation:

Given surds are of order 2 and 3, Their LCM is 6.

Changing each to a surd of order 6, we get :

√2=$2^{\dfrac{1}{2}}$

=$2^{\dfrac{1}{2}\times\dfrac{3}{3} }$

=$2^{\dfrac{3}{6}}$

=$(2^{3})^{\dfrac{1}{6}}$

=$(8)^{\dfrac{1}{6}}$

=$\sqrt[6]{8}$

$\sqrt[3]{3}=3^{\dfrac{1}{3}}$

=$3^{\dfrac{1}{3}\times\dfrac{2}{2} }$

=$3^{\dfrac{2}{6}}$

=$(3^{2})^{\dfrac{1}{6}}$

=$(9)^{\dfrac{1}{6}}$

=$\sqrt[6]{9}$

Clearly, $\sqrt[6]{9}$ > $\sqrt[6]{8}$ and hence $\sqrt[3]{3}$ > √2

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