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. 

$$\sqrt[3]{64}=4\;,\;\;\sqrt[3]{16}=4^{2/3}\;\implies$$
$$\sqrt[3]4+\sqrt[3]{16}+\sqrt[3]{64}=4^{1/3}+4^{2/3}+4=4^{1/3}\left(1+4^{1/3}+4^{2/3}\right)=$$
$$=4^{1/3}\frac{1-4}{1-4^{1/3}}=3\cdot4^{1/3}\frac1{4^{1/3}-1}\implies$$
$$\frac6{\sqrt[3]4+\sqrt[3]{16}+\sqrt[3]{64}}=\frac{2(4^{1/3}-1)}{4^{1/3}}=2^{1/3}(4^{1/3}-1)=2-\sqrt[3]2$$
