Question

Q13 Rewrite the following irrational numbers in descending order:
cube root of 10, cuberoot of 36,squareroot of 3 ,6root of 5 and 8th root of 60​

Answer 1

Answer:

$\sqrt[3]{36}$ , $\sqrt[3]{10}$ , √3 , $\sqrt[8]{60}$ , $\sqrt[6]{5}$

Step-by-step explanation:

$\sqrt[3]{10}$ , $\sqrt[3]{36}$ , √3 , $\sqrt[6]{5}$ , $\sqrt[8]{60}$

lcm of 3,2,6,8 = 24

so, $\sqrt[3]{10}$ , $\sqrt[3]{36}$ , √3 , $\sqrt[6]{5}$ , $\sqrt[8]{60}$

 = $\sqrt[3]{\sqrt[8]{10^{8} } }$ , $\sqrt[3]{\sqrt[8]{36^{8} } }$ , $\sqrt[2]{\sqrt[12]{3^{12} } }$ , $\sqrt[6]{\sqrt[4]{5^{4} } }$ , $\sqrt[8]{\sqrt[3]{60^3} }$

 = $\sqrt[24]{10^8}$ , $\sqrt[24]{36^8}$ , $\sqrt[24]{3^{12}}$ , $\sqrt[24]{5^4}$ , $\sqrt[24]{60^3}$

in descending order,

$\sqrt[24]{36^8}$ , $\sqrt[24]{10^8}$ ,  $\sqrt[24]{3^{12}}$ , $\sqrt[24]{60^3}$ , $\sqrt[24]{5^4}$

ie., $\sqrt[3]{36}$ , $\sqrt[3]{10}$ , √3 , $\sqrt[8]{60}$ , $\sqrt[6]{5}$