## Intermediate Algebra (6th Edition)

$x=-\frac{9}{2}$
$\sqrt[3] (6x)=-3$ Cube both sides of the equation. $(\sqrt[3] (6x))^{3}=(-3)^{3}$ $6x=-27$ Divide both sides by 6. $x=-\frac{27}{6}=-\frac{9}{2}$