jmc

Since $21\cdot 3=63=2\cdot 31+1$, it follows that $21$ is the modular inverse of $3$, modulo $31$. Thus, $2^n\equiv 2^{21}(\mathrm{mod}31)$. After computing some powers of $2$, we notice that $2^5\equiv 1(\mathrm{mod}31)$, so $2^{21}\equiv 2\cdot {\left(2^5\right)}^4\equiv 2(\mathrm{mod}31)$. Thus, ${\left(2^{21}\right)}^3\equiv 2^3\equiv 8(\mathrm{mod}31)$, and $${\left(2^{21}\right)}^3-2\equiv 8-2\equiv \boxed{6}(\mathrm{mod}31)$$Notice that this problem implies that ${\left(a^{3^{-1}}\right)}^3\not\equiv a(\mathrm{mod}p)$ in general, so that certain properties of modular inverses do not extend to exponentiation (for that, one needs to turn to Fermat's Little Theorem or other related theorems).