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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 = $-1/2$ + $\sqrt{3}i/2$

JCase
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    In the problem statement, $w$ is a complex number. So the real root times $w$ or $w^2$ is not real, but complex. – scott Mar 28 '17 at 21:35
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    I would start by writing $x^3=r$ and solving $x^3-r=0$ for $x$.Other way could be to prove that $(w\sqrt[3]{r})^3=r$ – kingW3 Mar 28 '17 at 21:36
  • cub roots of what? – fleablood Mar 28 '17 at 21:37
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    What is w and w^2. If w and w^2 are real numbers then $w\sqrt[3]r$ is a cube root of $w^3r$. And if $w$ is not real then $w\sqrt[3]r$ is only a real cube root of $w^3$ is real. Do you mean $w$ and $w$ are the complex cube roots of 1 (or -1)? – fleablood Mar 28 '17 at 21:40

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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 $1$?

fleablood
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