Question

The greatest among $\sqrt[6]{63}$, $\sqrt{2}$, $\sqrt[3]{34}$, $\sqrt[4]{45}$ is--

• A. $\sqrt[6]{63}$
• B. $\sqrt{2}$
• C. $\sqrt[3]{34}$
• D. $\sqrt[4]{45}$

Answer 1

Answer: (C)

$\sqrt[3]{34}$

$\therefore$ $\sqrt[6]{63}$ = ${\left(3\right)}^{\frac{1}{6}}$ = ${\left(3^2\right)}^{\frac{1}{12}}$ = ${\left(9\right)}^{\frac{1}{12}}$
$\sqrt{2}$ = ${\left(2\right)}^{\frac{1}{2}}$ = ${\left(2^6\right)}^{\frac{1}{12}}$ = ${\left(64\right)}^{\frac{1}{12}}$
$\sqrt[3]{34}$ = ${\left(4\right)}^{\frac{1}{3}}$ = ${\left(4^4\right)}^{\frac{1}{12}}$ = ${\left(256\right)}^{\frac{1}{12}}$

$\sqrt[4]{45}=(5{)}^{\frac{1}{4}}=(5^3{)}^{\frac{1}{12}}=(125{)}^{\frac{1}{12}}$
$\therefore$ The greatest number is ${\left(256\right)}^{\frac{1}{12}}$ = $\sqrt[3]{34}$