Rationalizing denominator with cube roots Rationalize the denominator of $$\frac{6}{\sqrt[3]{4}+\sqrt[3]{16}+\sqrt[3]{64}}$$ and simplify.
I already have an answer. I just want to compare answers with others. Maybe someone has different solutions? Also, I really disagree with the answer found at the back of the questionnaire. 

Use  $x+x^2+x^3=x\cdot(x^3-1)/(x-1)$ with $x=\sqrt[3]{4}$.
Hence 
$$\frac{6}{\sqrt[3]{4}+\sqrt[3]{16}+\sqrt[3]{64}}=\frac{6(\sqrt[3]{4}-1)}{\sqrt[3]{4}(4-1)}=2-2^{1-2/3}=2-\sqrt[3]{2}.$$
