## Precalculus: Mathematics for Calculus, 7th Edition

$a)$ $25$ $b)$ $\dfrac{125}{512}$ $c)$ $\dfrac{1}{81}$
$a)$ $125^{2/3}$ Rewrite this as a radical expression and evaluate: $125^{2/3}=\sqrt[3]{125^{2}}=\sqrt[3]{15625}=25$ $b)$ $\Big(\dfrac{25}{64}\Big)^{3/2}$ Rewrite this as a radical expression and evaluate: $\Big(\dfrac{25}{64}\Big)^{3/2}=\sqrt{\Big(\dfrac{25}{64}\Big)^{3}}=\Big(\dfrac{25}{64}\Big)\sqrt{\dfrac{25}{64}}=\Big(\dfrac{25}{64}\Big)\Big(\dfrac{5}{8}\Big)=\dfrac{125}{512}$ $c)$ $27^{-4/3}$ Rewrite the expression to change the sign of the exponent: $27^{-4/3}=\dfrac{1}{27^{4/3}}=...$ Now, rewrite the denominator as a radical expression and evaluate: $...=\dfrac{1}{\sqrt[3]{27^{4}}}=\dfrac{1}{27\sqrt[3]{27}}=\dfrac{1}{27(3)}=\dfrac{1}{81}$