## Problem 1

${\displaystyle {\sqrt {25}}}$ or ${\displaystyle 25^{\left({\frac {1}{2}}\right)}}$

5

## Problem 2

${\displaystyle 27^{\left({\frac {1}{3}}\right)}}$ or ${\displaystyle {\sqrt[{3}]{27}}}$

${\displaystyle (27)^{\left({\frac {1}{3}}\right)}}$ = 3 because (3*3*3) or ${\displaystyle 3^{3}}$ = 27 then the cube root of 27 is 3.

## Problem 3

${\displaystyle {\sqrt[{4}]{648+648}}+8}$ or ${\displaystyle (648+648)^{\frac {1}{4}}+8}$

${\displaystyle {\sqrt[{4}]{1296}}+8=6+8=14}$
${\displaystyle 512^{\frac {2}{3}}}$ or ${\displaystyle ({\sqrt[{3}]{512}})^{2}}$
${\displaystyle (512)^{\frac {2}{3}}=(8)^{2}=64}$