#### Answer

$x^{1/15}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
Use the laws of exponents to simplify the given expression, $
x^{2/5}\cdot x^{-1/3}
.$
$\bf{\text{Solution Details:}}$
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}
x^{\frac{2}{5}+\left(-\frac{1}{3}\right)}
\\\\=
x^{\frac{2}{5}-\frac{1}{3}}
.\end{array}
To simplify the expression, $
\dfrac{2}{5}-\dfrac{1}{3}
,$ find the $LCD$ of the denominators $\left\{
5,3
\right\}.$ The $LCD$ is $
15
$ since it is the lowest number that can be exactly divided by the denominators. Multiplying both the numerator and denominator of each terms by an expression equal to $1$ which will make the denominators equal to the $LCD$ results to
\begin{array}{l}\require{cancel}
x^{\frac{2}{5}\cdot\frac{3}{3}-\frac{1}{3}\cdot\frac{5}{5}}
\\\\=
x^{\frac{6}{15}-\frac{5}{15}}
\\\\=
x^{\frac{1}{15}}
\\\\=
x^{1/15}
.\end{array}