#### Answer

$k^{2/3}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
Use the laws of exponents to simplify the given expression, $
\dfrac{k^{1/3}}{k^{2/3}\cdot k^{-1}}
.$
$\bf{\text{Solution Details:}}$
Using the Product Rule of the laws of exponents which is given by $x^m\cdot x^n=x^{m+n},$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{k^{1/3}}{k^{\frac{2}{3}+(-1)}}
\\\\=
\dfrac{k^{1/3}}{k^{\frac{2}{3}-1}}
.\end{array}
Using the Quotient Rule of the laws of exponents which states that $\dfrac{x^m}{x^n}=x^{m-n},$ the expression above simplifies to
\begin{array}{l}\require{cancel}
k^{\frac{1}{3}-\left( \frac{2}{3}-1 \right)}
\\\\=
k^{\frac{1}{3}-\frac{2}{3}+1}
.\end{array}
To simplify the expression $
\dfrac{1}{3}-\dfrac{2}{3}+1
,$ change the expressions to similar fractions (same denominator) by using the $LCD$. The $LCD$ of the denominators $
3,3,
$ and $
1
$ is $
3
$ since it is the lowest number that can be exactly divided by the denominators. Multiplying the terms by an expression equal to $1$ that will make the denominator equal to the $LCD$ results to
\begin{array}{l}\require{cancel}
k^{\frac{1}{3}-\frac{2}{3}+1\cdot\frac{3}{3}}
\\\\=
k^{\frac{1}{3}-\frac{2}{3}+\frac{3}{3}}
\\\\=
k^{\frac{2}{3}}
\\\\=
k^{2/3}
.\end{array}