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How am I supposed to rationalize the denominator for $\frac {1}{\sqrt[3]{x}+2}$?

I don't even know where to really begin. I tried multiplying both the numerator and denominator by $\sqrt[3]{x}-2$, but that didn't work at all...

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    $\begingroup$ HINT: $(a+b)(a^2-ab+b^2)=a^3+b^3$ $\endgroup$ – Brian M. Scott May 1 '16 at 19:22
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You will have to multiply both the numerator and the denominator by sums of cubes.

Sums of Cubes: $a^3+b^3=(a+b)(a^2-ab+b^2)$

So to get $\sqrt[3]{x}+2$ into a perfect number with no radicals, you have to multiply it by $\sqrt[3]{x^2}-2\sqrt[3]{x}+\sqrt[3]{64}$. Doing so to the numerator and denominator gives us $$\boxed{\frac {\sqrt[3]{x^2}-2\sqrt[3]{x}+4}{x+8}}$$

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