## Precalculus (6th Edition)

$x=4$
Cube both sides to obtain: $$(\sqrt[3]{2x+1})^3=(\sqrt[3]{9})^3 \\2x+1=9$$ Subtract $1$ to both sides: $$2x+1-1=9-1 \\2x=8$$ Divide $2$ to both sides: $$\dfrac{2x}{2}=\dfrac{8}{2} \\x=4$$