Answer
$\pm2$
Work Step by Step
The given values are
$n=6$ and $a=64$
$n$ is even and $a>0$, then $a$ has two real $n^{th}$ roots.
$=\pm\sqrt[n]{a}$
Substitute the values of $n$ and $a$.
$=\pm\sqrt[6]{64}$
Use $64=(2)^6$
$=\pm\sqrt[6]{2^6}$
Simplify.
$=\pm2$.
