## Algebra 1: Common Core (15th Edition)

$c^\frac{1}{3}$
$(c^3)^\frac{1}{9}$ $\times$ $(d^3)^0$ You have to multiply exponents when raising a power to a power. So we get: $c^{(3\times\frac{1}{9})}$$\times d^{(3\times0)} =c^\frac{3}{9}$$\times$ $d^0$ =$c^\frac{1}{3}$ $\times 1$ = $c^\frac{1}{3}$