Question

the greatest no among cube root of 9 , fourth root of 11, and sixth root of 17 ​

Answer 1

$\huge\mathfrak{Answer}$

∛9

$\huge\textsf{Explanation}$

We have to find greatest number among :-

$\sf \sqrt[3]{9} ,\sqrt[4]{11} \ and \sqrt[6]{17} \\\\ \sqrt[3]{9} \ can\ be\ written\ as\ 9^{\frac{1}{3}}\\\\Similarly, \sqrt[4]{11} \ and \sqrt[6]{17} \ would\ be\ 11^{\frac{1}{4}}\ and 17^{\frac{1}{6}}\\\\Now, take \ the\ LCM \ of \ the\ powers\\ and\ make\ their\ denominators\ same \\\\9^{\frac{1}{3}} = 9^\frac{4}{12}\\\\11^\frac{1}{4}= 11^\frac{3}{12}\\\\17^\frac{1}{6} = 17^\frac{2}{12}\\\\Now, just\ compare\ the\ numbers\ by\ solving\ them\\ with \\numerators$$\sf so, 9^\frac{4}{12} = 6561^\frac{1}{12}\\\\11^\frac{3}{12} = 1331^\frac{1}{12}\\\\17^\frac{2}{12}= 289^\frac{1}{12}\\\\\\here, 6561\ is \ the \ largest \ number \ hence, \\we \ say \ that \ 9^\frac{1}{3} \ is \ the \greatest\ number$

$\rule{200}2$

Answer 2

Answer:

$\sqrt[3]{9}$