## Precalculus (6th Edition)

$\color{blue}{-\dfrac{4}{3}}$
RECALL: (1) $a^{1/n} = \sqrt[n]{a}$ (2) If $a \gt 0$, then $\sqrt[n]{a^n} = a$ (3) If $n$ is odd, then $\sqrt[n]{a^n}=a$ Use rule (1) above to obtain: $\left(-\dfrac{64}{27}\right)^{1/3} = \sqrt[3]{-\left(\dfrac{64}{27}\right)}$ Since $-\dfrac{64}{27}=\left(-\dfrac{4}{3}\right)^3$, then the expression above is equivalent to: $=\sqrt[3]{\left(-\dfrac{4}{3}\right)^3}$ Use rule (3) above to simplify and have: $\sqrt[3]{\left(-\dfrac{4}{3}\right)^3}=\color{blue}{-\dfrac{4}{3}}$