Answer
$\color{blue}{k^{2/3}}$
Work Step by Step
RECALL:
(1) $a^m \cdot a^n = a^{m+n}$
(2) $\dfrac{a^m}{a^n} = a^{m-n}$
(3) $a^{1/n} = \sqrt[n]{a}$
(4) When $n$ is odd, $\sqrt[n]{a^n}=a$
(5) $a^{m/n} = \left(\sqrt[n]{a}\right)^m$
Use rule (1) above to obtain:
$=\dfrac{k^{1/3}}{k^{2/3+(-1)}}
\\=\dfrac{k^{1/3}}{k^{2/3+(-3/3)}}
\\=\dfrac{k^{1/3}}{k^{-1/3}}$
Use rule (2) above to obtain:
$=k^{1/3 - (-1/3)}
\\=k^{1/3+1/3}
\\=\color{blue}{k^{2/3}}$
