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.