How do I calculate $\lim \limits_{x \to 8} (\sqrt[3]x-2/(x-8))=1$ using the factorization formula?

I am asked to calculate the $ \lim \limits_{x \to 8} (\sqrt[3]x-2/(x-8))=1$ using the factorization formula

$ x^n-a^n = (x-a)(x^{n-1}+x^{n-2}a+x^{n-3}a^2+…+xa^{n-2}+a^{n-1})$

I have rewritten the limit as

$ \lim \limits_{x \to 8} (x^{1/3}-8^{1/3}/(x-8))=1$

I know that $ x-8$ will cancel out, however I do not know how to plug in the values of $ x$ and $ a$ into the formula. I do not know where to stop plugging in values, the fractional exponent is confusing me.