We determine $p$ such that $(\sqrt[7]{x^2})^{-3}\cdot x^5=x^p$. We have the following steps
$$\begin{align}
(\sqrt[7]{x^2})^{-3}\cdot x^5 & = \frac{1}{(\sqrt[7]{x^2})^{3}}\cdot x^5\\
& = \frac{1}{(x^{\frac{2}{7}})^{3}}\cdot x^5\\
& = \frac{1}{x^{\frac{6}{7}}}\cdot x^5\\
& = \frac{x^5}{x^{\frac{6}{7}}}\\
&= x^{4\frac{1}{7}},
\end{align}$$
which results in $p=4\frac{1}{7}$.
