Limits problem: Factoring a cube root of x?

Disclaimer: I am an adult learning Calculus. This is not a student posting his homework assignment. I think this is a great forum!

$$\lim_{x\to8}{\frac{\sqrt[3] x-2}{x-8}}$$

How do I factor the top to cancel the $x-8$ denominator?

Answer

Actually you need to factor the denominator: $x-8 = (\sqrt[3]{x})^3-2^3 = (\sqrt[3]{x}-2)\left((\sqrt[3]{x})^2+2\sqrt[3]{x}+2^2\right)$.