how to rationalize $\frac {x-8}{\sqrt[3]{x}-2}$

Question

In order to resolve a limit, I need to rationalize $\frac {x-8}{\sqrt[3]{x}-2}$. I tried multiplying it by $\sqrt[3]{x^3}$ or $\sqrt[3]{x^2}$ but with no much success. It seems that I can't use "Difference of Squares" identity too. The limit in question is:

$$ \textstyle \lim_{x \to 8}\frac {x-8}{\sqrt[3]{x}-2} $$

When ${x \to 8}$, the limit is indeterminate. I know there is l'Hôpital's rule, but I can't use it now, because I have not read the topic of derivatives yet (I have started from the beginning).

Answer 1

Hint: Let $x^{1/3}=y$. Then we are looking at $\dfrac{y^3-8}{y-2}$.

Answer 2

Notice, $(a-b)(a^2+b^2+ab)=a^3-b^3$, now we have $$\frac{x-8}{x^{1/3}-2}$$ $$=\frac{(x-8)(x^{2/3}+4+2x^{1/3})}{(x^{1/3}-2)(x^{2/3}+4+2x^{1/3})}$$ $$=\frac{(x-8)(x^{2/3}+4+2x^{1/3})}{((x^{1/3})^3-(2)^3)}$$ $$=\frac{(x-8)(x^{2/3}+4+2x^{1/3})}{(x-8)}$$ $$=\color{blue}{x^{2/3}+2x^{1/3}+4}$$