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. 

Write $x=\sqrt[3]4$, then the expression is
$$\frac6{x+x^2+x^3}=\frac1x×\frac6{1+x+x^2}$$
Multiply by $x-1$:
$$=\frac1x×\frac{6(x-1)}{(x-1)(x^2+x+1)}$$
$$=\frac1x×\frac{6(x-1)}{x^3-1}$$
$$=\frac{x-1}x×\frac{6}{3}$$
$$=2\left(1-\frac1x\right)$$
$$=2\left(1-\frac{x^2}{x^3}\right)$$
$$=2\left(1-\frac{\sqrt[3]{16}}4\right)$$
$$=2\left(1-\frac{2\sqrt[3]{2}}4\right)$$
$$=2-\sqrt[3]2$$
