## Intermediate Algebra (6th Edition)

We are asked to evaluate $\sqrt 64-\sqrt[3] 64$. We know that $\sqrt 64=8$, because $8^{2}=8\times8=64$ And we know that $\sqrt[3] 64=4$, because $4^{3}=4\times4\times4=64$ Therefore, $\sqrt 64-\sqrt[3] 64=8-4=4$