Question

pls solve this with steps

Answer 1

3 underroot 3 is greater

Answer 2

Lcm of 3,4 = 12.

$= \ \textgreater \ Given : \sqrt[3]{3}$

$= \ \textgreater \ (3^ \frac{1}{3} )^{12}$

$= \ \textgreater \ 3^{ \frac{12}{3} }$

$= \ \textgreater \ 3^4$

81.


Now,

$Given : \sqrt[4]{4}$

$= \ \textgreater \ (4^ \frac{1}{4} )^{ \frac{1}{12} }$

$= \ \textgreater \ 4^{ \frac{12}{4} }$

$= \ \textgreater \ 4^3$

64.


Therefore $\sqrt[3]{3}$ is greater.


Hope this helps!