Question

The greatest number among $\sqrt[3]{39}$, $\sqrt[4]{411}$, $\sqrt[6]{617}$, $\sqrt[8]{881}$.

• A. $\sqrt[3]{39}$.
• B. $\sqrt[4]{411}$
• C. $\sqrt[6]{617}$
• D. $\sqrt[8]{881}$

Answer 1

The correct option is A $\sqrt[3]{39}$.

$\sqrt[3]{39}$, $\sqrt[4]{411}$, $\sqrt[6]{617}$, $\sqrt[8]{881}$

LCM of 3, 4, 6 and 8 is 24.

$\sqrt[3]{39}$ = $9^{\frac{1}{3}}$ = $({9^{\frac{1}{3}})}^{24}$ = $9^{\frac{1}{3}\times 24}$ = $9^8$

$\sqrt[4]{411}$ = ${11}^{\frac{1}{4}}$ = $({{11}^{\frac{1}{4}})}^{24}$ = ${11}^{\frac{1}{4}\times 24}$ = ${11}^6$

$\sqrt[6]{617}$ = ${17}^{\frac{1}{6}}$ = $({{17}^{\frac{1}{6}})}^{24}$ = ${17}^{\frac{1}{6}\times 24}$ = ${17}^4$

$\sqrt[8]{881}$ = ${81}^{\frac{1}{8}}$ = $({{81}^{\frac{1}{8}})}^{24}$ = ${81}^{\frac{1}{8}\times 24}$ = ${81}^3$

Now, we can write
$9^8$ =${\left(9^2\right)}^4$= ${81}^4$
So, ${81}^4$ is greater than ${17}^4$ and ${81}^3$

To compare ${81}^4$ and ${11}^6$, we can take the ratio of the given two numbers;
${81}^4$ : ${11}^6$, using approximation the given ratio is greater than 1.

Hence, $\sqrt[3]{39}$ is the greatest among the given numbers.