## Precalculus (6th Edition)

$\color{blue}{\dfrac{1}{x^{10/3}}}$
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) $a^{m/n} = \left(\sqrt[n]{a}\right)^m$ (6) $(ab)^m=a^mb^m$ (7) $(a^m)^n=a^{mn}$ (8) $a^{-m}=\dfrac{1}{a^m}$ Use rule (7) above to obtain: $=\dfrac{x^{(2/3) \cdot 2}}{x^{2 \cdot (7/3)}} \\=\dfrac{x^{4/3}}{x^{14/3}}$ Use rule (2) above to obtain: $=x^{4/3-14/3} \\=x^{-10/3}$ Use rule (8) above to obtain: $=\color{blue}{\dfrac{1}{x^{10/3}}}$