#### Answer

the domain is: $\{x|x\neq-4\}$

#### Work Step by Step

We are given:
$f(x)= \frac{\sqrt[3]{2x+1}}{\sqrt[3]{2x}+2}$
We know that the denominator can not be 0:
$\sqrt[3]{2x}+2\neq 0$
$\sqrt[3]{2x}\neq-2$
$2x\neq(-2)^3$
$2x\neq-8$
$x\neq-4$
Therefore, the domain is: $\{x|x\neq-4\}$.