Example (film) | Wiskunde op Tilburg University

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

$p=4\frac{1}{7}$ .