Find the value of 27^(4/3), 243^(2/5) and 625^(-1/4).

$a^m a^n = a^{m+n}$, $(a^m)^n = a^{mn}$, $\dfrac{a^m}{a^n} = a^{m-n}$, $a^0 = 1$ and $a^{-n} = \dfrac{1}{a^n}$. $a^{1/n}$ is the $n$-th root of $a$, and $a^{m/n} = \left(\sqrt[n]{a}\right)^m$: take the root first.