Evaluating $\lim\limits_{x\to 2} \frac{\sqrt[3]{x} - \sqrt[3]{2}}{x^3 - 8}$ without l'Hospital's rule 
How to evaluate the following limit?
  $$
\lim\limits_{x\to 2} \dfrac{\sqrt[3]{x} - \sqrt[3]{2}}{x^3 - 8}
$$

I factored the denominator into $(x-2)(x^2+2x+4)$, but I couldn't go on from there.

Hint:
It's $\lim_\limits{y\to\sqrt[3]2}\dfrac{y-\sqrt[3]2}{y^9-8}=\lim_\limits{y\to\sqrt[3]2}\dfrac{y-\sqrt[3]2}{(y^3-2)(y^6+2y^3+4)}$
$$=\lim_\limits{y\to\sqrt[3]2}\dfrac{y-\sqrt[3]2}{(y-\sqrt[3]2)(y^2+\sqrt[3]2y+\sqrt[3]2^{2})(y^6+2y^3+4)}$$
