If r is any real-number, then r has a real cube root, which is $\sqrt[3]{r}$. Show that $w$$\sqrt[3]{r}$ and $w^2$$\sqrt[3]{r}$ are also cube roots.
Not really sure how to answer this. I know that cubes have a real root and a couple imaginary. So I guess the real root times $w$ or $w^2$ would still be real.
EDIT: w = $-1/2$ + $\sqrt{3}i/2$
Well, um, $(w^k\sqrt[3]r)^3 = w^{3k}r = (w^3)r=r$. I'm assuming you meant $w$ is complex root of $1$?