Simplify $ \frac{ \sqrt[3]{16} - 1}{ \sqrt[3]{27} + \sqrt[3]{4} + \sqrt[3]{2}} $ Simplify
$$ \frac{ \sqrt[3]{16} - 1}{ \sqrt[3]{27} + \sqrt[3]{4} + \sqrt[3]{2}} $$

Attempt:
$$ \frac{ \sqrt[3]{16} - 1}{3 + \sqrt[3]{4} + \sqrt[3]{2}} =  \frac{ \sqrt[3]{16} - 1}{ (3 + \sqrt[3]{4}) + \sqrt[3]{2}} \times \frac{ (3 + \sqrt[3]{4}) - \sqrt[3]{2}}{  (3 + \sqrt[3]{4}) - \sqrt[3]{2}}  $$
$$ =    \frac{ (\sqrt[3]{16} - 1) [(3 + \sqrt[3]{4}) - \sqrt[3]{2}]}{ (3 + \sqrt[3]{4})^{2} - 2^{2/3}}  $$
$$ =     \frac{ 3 \sqrt[3]{16} - 3\sqrt[3]{4} + \sqrt[3]{2} + 1}{ (9 + 5 \sqrt[3]{4} + \sqrt[3]{16}) } $$
From here on I don't know how to continue.
I can let $a = \sqrt[3]{2}$, but still cannot do anything.

With the same notation as in the other answer, i.e. $x = \sqrt[3] 2$, noting that $x^3+1=3,$ you can write your quantity as
\begin{eqnarray}
\frac{x^4-1}{x^2+x+3} &=& \frac{(x-1)(x^3+x^2+x+1)}{x^2+x+x^3+1}=\\
&=&x-1.
\end{eqnarray}
