## Intermediate Algebra (6th Edition)

$-\sqrt[3] (x^{3})=-(x)=-x$ We know that $\sqrt[3] (x^{3})=x$ because $(x^{3})^\frac{1}{3}=(x)^{3\times\frac{1}{3}}=x$ In this case, we are taking the opposite of $\sqrt[3] (x^{3})$. We do not need absolute value bars around the result, because the index of the radical is odd in this case.