Simplifying $\frac{3 - 8}{\sqrt[3]{9} + 2\sqrt[3]{3} + 4} - \sqrt[3]{3}$. How to find the value? 
Find the meaning of the expression
$$\frac{3 - 8}{\sqrt[3]{9} + 2\sqrt[3]{3} + 4} - \sqrt[3]{3}$$

What I tried:
$$\tag{1} \frac{-5}{\sqrt[3]{3^2} + 2\sqrt[3]{3} + 2^2} - \frac{(\sqrt[3]{3})(\sqrt[3]{3^2}) + (\sqrt[3]{3})(2\sqrt[3]{3}) + 4(\sqrt[3]{3})}{\sqrt[3]{3^2} + 2\sqrt[3]{3} + 2^2}$$
$$\tag{2} =\frac{-5}{\sqrt[3]{3^2} + 2\sqrt[3]{3} + 2^2} - \frac{3 + 6 + 4(\sqrt[3]{3})}{\sqrt[3]{3^2} + 2\sqrt[3]{3} + 2^2}$$
$$\tag{3} =\frac{-5 - 3 - 6 - 4(\sqrt[3]{3})}{\sqrt[3]{3^2} + 2\sqrt[3]{3} + 2^2} $$
$$\tag{4} =\frac{-14 - 4(\sqrt[3]{3})}{\sqrt[3]{3^2} + 2\sqrt[3]{3} + 2^2}$$

Note that in your computation from (1) to (2), it should be
$$(\sqrt[3]{3})(2\sqrt[3]{3})=2\sqrt[3]{3^2}\qquad (\text{it is not $6$)}.$$
It easier to follow the algebraic manipulations if we let $a=\sqrt[3]{3}$:
$$\begin{align}\frac{3 - 8}{\sqrt[3]{9} + 2\sqrt[3]{3} + 4} - \sqrt[3]{3}&=\frac{a^3-8}{a^2+2a+4}-a=\frac{a^3-8-a(a^2+2a+4)}{a^2+2a+4}\\&=
-\frac{2(a^2+2a+4)}{a^2+2a+4}=-2.\end{align}$$
