Harvard-MIT Mathematics Tournament

Solution:
Answer: $\frac{5}{3}$

Make the substitution $t=\frac{1}{x}$. Then the limit equals to
$$\lim_{t \rightarrow 0} t^{-p}\left(\sqrt[3]{\frac{1}{t}+1}+\sqrt[3]{\frac{1}{t}-1}-2 \sqrt[3]{\frac{1}{t}}\right)=\lim_{t \rightarrow 0} t^{-p-\frac{1}{3}}(\sqrt[3]{1+t}+\sqrt[3]{1-t}-2)$$
We need the degree of the first nonzero term in the Maclaurin expansion of $\sqrt[3]{1+t}+\sqrt[3]{1-t}-2$. We have
$$\sqrt[3]{1+t}=1+\frac{1}{3} t-\frac{1}{9} t^{2}+o\left(t^{2}\right), \quad \sqrt[3]{1-t}=1-\frac{1}{3} t-\frac{1}{9} t^{2}+o\left(t^{2}\right)$$
It follows that $\sqrt[3]{1+t}+\sqrt[3]{1-t}-2=-\frac{2}{9} t^{2}+o\left(t^{2}\right)$. By considering the degree of the leading term, it follows that $-p-\frac{1}{3}=-2$. So $p=\frac{5}{3}$.